NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

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Q.1. Examine the continuity of the function 
f(x) = x3 + 2x2 – 1 at x = 1
Ans.
We know that y = f(x) will be continuous at x = a if
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Given: f(x) = x3 + 2x– 1
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence, f(x) is continuous at x = 1.

Find which of  the functions in Exercises 2 to 10 is continuous or discontinuous at the indicated points:
Q.2. NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRevat x = 2
Ans.
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
SinceNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence f (x) is discontinuous at x = 2.

Q.3. NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRevat x = 0
Ans.
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
AsNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
∴ f(x) is discontinuous at x = 0.

Q.4. NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRevat x = 2
Ans.
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence, f(x) is continuous at x = 2.

Q.5. NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRevat x = 4
Ans.
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRevNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence, f(x) is discontinuous at x = 4.

Q.6. NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRevat x = 0
Ans.
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence, f(x) is continuous at x = 0.

Q.7. NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRevat x = a
Ans.
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
= 0 × [a number oscillating between – 1 and 1]
= 0
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
= 0 × [a number oscillating between – 1 and 1]
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
AsNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence, f(x) is continuous at x = a.

Q.8. NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRevat x = 0
Ans.
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev[∵ e = 0]
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev[e– ∞= 0]
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
AsNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence, f(x) is discontinuous at x = 0.

Q.9. NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRevat x = 1
Ans.
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence, f(x) is continuous at x = 1.

Q.10. NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Ans.
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence, f(x) is continuous at x = 1.

Find the value of k in each of the Exercises 11 to 14 so that the function f is continuous at the indicated point:
Q.11. NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRevat x = 5
Ans.
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
As the function is continuous at x = 5
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence, the value of k isNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.12. NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRevat x = 2
Ans.
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
As the function is continuous at x = 2.
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence, value of k isNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.13. NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRevat x = 0
Ans.
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
As the function is continuous at x = 0.
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
k = – 1
Hence, the value of k is – 1.

Q.14. NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRevat x = 0
Ans.
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRevNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
⇒ k2 = 1 ⇒ k = ± 1
Hence, the value of k is ± 1.

Q.15. Prove that the function f defined by
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
remains discontinuous at x = 0, regardless the choice of k.
Ans.
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence, f(x) is discontinuous at x = 0 regardless the choice of k.

Q.16. Find the values of a and b such that the function f  defined by
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
is a continuous function at x = 4.
Ans.
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
As the function is continuous at x = 4.
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
– 1 + a = a + b = 1 + b
 – 1 + a=a + b ⇒ b = – 1
1 + b =a + b ⇒ a = 1
Hence, the value of a = 1 and b = – 1.

Q.17. Given the function f (x) =NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRevFind the points of discontinuity of the composite function y = f (f (x)).
Ans.
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
This function will not be defined and continuous where
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev    
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRevis the point of discontinuity.

Q.18 Find all points of discontinuity of the functionNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
whereNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Ans.
We haveNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRevNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
So, if f(t) is discontinuous, then 2 – x = 0
∴ x = 2 and 2x – 1 = 0NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence, the required points of discontinuity are 2 and 1/2.

Q.19. Show that the function f(x) = sin x + cos x is continuous at x = π.
Ans.
Given that f(x) =NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Put g(x) = sin x + cos x and h(x) =NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
∴ h[g(x)] = h(sin x + cos x) =NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Now, g(x) = sin x + cos x is a continuous function since sin x and cos x are two continuous functions at x = π.

We know that every modulus function is a continuous function everywhere.

Hence, f(x) =NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRevis continuous function at x = π.

Examine the differentiability of f, where f is defined by

Q.20. NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRevat x = 2.
Ans.
We know that a function f is differentiable at a point ‘a’ in its domain if
Lf ′(c) = Rf ′(c)
where Lf ′(c) =NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRevNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Lf ′(2) ≠ Rf ′(2)
Hence, f(x) is not differentiable at x = 2.

Q.21. Examine the differentiability of f, where f is defined by
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRevat x = 0.
Ans.
Given that:
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
For differentiability we know that:′
Lf ′(c) = Rf ′(c)
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
So, Lf ′(0) = Rf ′(0)  = 0
Hence, f(x) is differentiable at x = 0.

Q.22. NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRevat x = 2.
Ans.
 f(x) is differentiable at x = 2 if
Lf ′(2) = Rf ′(2)
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
So, Lf ′(2) ≠ Rf ′(2)
Hence, f(x) is not differentiable at x = 2.

Q.23. Show that f(x) = NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev is continuous but not differentiable at x = 5.
Ans.
We have f(x) =NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
For continuity at x = 5
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
L.H.L. = R.H.L.
So, f(x) is continuous at x = 5.
Now, for differentiability
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
∵ Lf ′(5) ≠ Rf ′(5)
Hence, f(x) is not differentiable at x = 5.

Q.24. A function f : R → R satisfies the equation f( x + y) = f(x) f(y) for all x, y ∈ R, f(x) ≠ 0. Suppose that the function is differentiable at x = 0 and f ′ (0) = 2. Prove that f ′(x) = 2 f (x).
Ans.
Given that: f : R → R satisfies the equation f(x + y) = f(x).f(y) ∀ x, y ∈ R, f(x) ≠ 0.
Let us take any point x = 0 at which the function f(x) is differentiable.
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NowNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence, f′(x) = 2f(x).

Differentiate each of the following w.r.t. x (Exercises 25 to 43) :
Q.25.NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Ans.
LetNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Taking log on both sides, we get
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Differentiating both sides w.r.t. x
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.26.NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Ans.
LetNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Taking log on both sides, we get,  log y =NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
⇒ log y = log 8x – log x8 ⇒ log y = x log 8 – 8 log x
Differentiating both sides w.r.t. x
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.27.NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Ans.
LetNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Differentiating both sides w.r.t. x
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.28. log [log (log x5 )]
Ans.
Let y = log [log (log x5 )]
Differentiating both sides w.r.t. x
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.29. sin √x + cos2 √x
Ans.
Let y = sin √x + cos2 √x
Differentiating both sides w.r.t. x
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.30. sinn (ax2 + bx + c)
Ans.
Let y = sinn (ax2 + bx + c)
Differentiating both sides w.r.t. x
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
= n.sinn – 1(ax2 + bx + c).cos(ax2 + bx + c).(2ax + b)
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.31. NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev 
Ans.
Let y =NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev 
Differentiating both sides w.r.t. x
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.32. sinx2 + sin2x + sin2(x2)
Ans.
Let y = sinx2 + sin2x + sin2(x2)
Differentiating both sides w.r.t. x,
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.33.NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev 

Ans.
LetNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev 
Differentiating both sides w.r.t. x
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.34. ( sin x )cos x
Ans.
Let y = (sin x)cos x
Taking log on both sides,
log y = log (sin x)cos x
⇒ log y = cos x.log (sin x) [∵ log xy = y log x]
Differentiating both sides w.r.t. x,
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.35. sinmx . cosnx
Ans.
 Let y = sinm x . cosn x
Differentiating both sides w.r.t. x
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
= n.sinm x.cosn – 1 x.(– sin x) + m.cosn x.sinm –1 x.cos x
= – n.sinm + 1 x.cosn – 1 x + m cosn + 1 x. sinm – 1x
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.36. (x + 1)2 (x + 2)3 (x + 3)4
Ans.
Let y = (x + 1)2(x + 2)3(x + 3)4 
Taking log on both sides,
log y = log [( x + 1)2 .(x+ 2)3 .( x + 3)4 ]
⇒ log y = log (x + 1)2 + log (x + 2)3 + log (x + 3)4
[∵ log xy = log x + log y]
⇒ log y = 2 log (x + 1) + 3 log (x + 2) + 4 log (x + 3)
[∵ log xy = y log x]
Differentiating both sides w.r.t. x,
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
= (x + 1)(x + 2)2(x + 3)3(2x2 + 10x + 12 + 3x2 + 12x + 9 + 4 x2 + 12x + 8)
= (x + 1)(x + 2)2(x + 3)3(9x2 + 34x + 29)
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.37.NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Ans.
LetNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRevNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Differentiating both sides w.r.t. x
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.38.NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Ans.
LetNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Differentiating both sides w.r.t. x
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.39. NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Ans.
Let y = tan– 1(sec x + tan x)
Differentiating both sides w.r.t. x
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Alternate solution
Let y =NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev 
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev[Dividing the Nr. and Den. by cos x/2]⇒
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Differentiating both sides w.r.t. x
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.40.NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Ans.
LetNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev 
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Differentiating both sides with respect to x
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.41. NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Ans.
LetNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Put x = cos θ ∴ θ = cos– 1 x
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
⇒ y = sec– 1 (sec 3θ) ⇒ y = 3θ
y = 3 cos– 1 x
Differentiating both sides w.r.t. x
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRevNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.42.NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Ans.
LetNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Put x = a tan θNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
⇒ y = tan - 1[tan 3θ]NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Differentiating both sides w.r.t. x
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.43.NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Ans.
LetNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Putting x2 = cos 2θNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Differentiating both sides w.r.t. x
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev   

FindNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev of each of the functions expressed in parametric form in Exercises from 44 to 48.
Q.44.NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Ans.
Given that:
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev 
Differentiating both the given parametric functions w.r.t. t
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.45.NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Ans.
Given that:
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev 
Differentiating both the parametric functions w.r.t. θ.
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.46. x = 3cosθ – 2cos3θ, y = 3sinθ – 2sin3θ.
Ans.
Given that: x = 3cosθ – 2cos3θ, y = 3sinθ – 2sin3θ.
Differentiating both the parametric functions w.r.t. θ
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
= – 3 sin θ – 6 cos2 θ . (– sin θ) 
= – 3 sin θ + 6 cos2 θ . sin θ
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
= 3 cos θ – 6 sin2 θ . cos θ
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.47. NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Ans.
Given that sin x =NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRevNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
 Taking sin x =NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Differentiating both sides w.r.t t, we get
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Now taking, tan y =NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Differentiating both sides w.r.t, t, we get
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.48. NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Ans.
Given that:NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Differentiating both the parametric functions w.r.t. t
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.49. If x = ecos2t and y = esin2t, prove thatNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Ans.
Given that: x = ecos2t and y = esin2t 
⇒ cos 2t = log x and sin 2t = log y.
Differentiating both the parametric functions w.r.t. t
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Now y = esin2t
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.50. If x = a sin2t (1 + cos2t) and y = b cos2t (1–cos2t), show that
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Ans.
Given that: x = a sin 2t (1 + cos 2t) and y = b cos 2t (1 – cos 2t).
Differentiating both the parametric functions w.r.t. t
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
= a [sin 2t . ( - sin 2t).2 + (1+ cos 2t) (cos 2t).2]
= a[- 2 sin 2 2t + 2 cos 2t+ 2 cos2 2t]
= a[2(cos2 2t - sin2 2t)+ 2 cos 2t]
= a [2 cos 4t + 2 cos 2t] [∵ cos 2x = cos2 x - sin2 x]
= 2a [cos 4t + cos 2t]
y = b cos 2t (1 - cos 2t)
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
= b [cos 2t. sin 2t.2 + (1- cos 2t ).( - sin 2t ).2]
= b [2 sin 2t. cos 2t - 2 sin 2t+ 2 sin 2t cos 2t]
= b [sin 4t - 2 sin 2t+ sin 4t] [∵ sin 2x = 2 sin x cos x]
= b [2 sin 4t - 2 sin 2t ] = 2b (sin 4t – sin 2t)
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
PutNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.51. If x = 3sint – sin 3t, y = 3cost – cos 3t, findNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Ans.
Given that: x = 3 sin t – sin 3t, y = 3 cos t – cos 3t.
Differentiating both parametric functions w.r.t. t
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev= 3 cos t – cos 3t.3 = 3(cos t – cos 3t)
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev= – 3 sin t + sin 3t.3 = 3(– sin t + sin 3t)
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
PutNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.52. DifferentiateNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRevw.r.t. sinx.
Ans.
LetNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRevand z = sin x.
Differentiating both the parametric functions w.r.t. x,
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.53. Differentiate tan–1NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRevw.r.t. tan–1 x when x ≠ 0.
Ans.
LetNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Put x = tan θ.
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Differentiating both parametric functions w.r.t. θ
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

FindNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev  when x and y are connected by the relation given in each of the Exercises 54 to 57.
Q.54.NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Ans.
Given that:NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Differentiating both sides w.r.t. x
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.55. sec (x + y) = xy
Ans.
Given that: sec (x + y) = xy
Differentiating both sides w.r.t. x
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.56. tan–1 (x2 + y2) = a
Ans.
Given that: tan– 1(x2 + y2) = a
⇒ x2 + y2 = tan a.
Differentiating both sides w.r.t. x.
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.57. (x2 + y2)2 = xy
Ans.
Given that: (x2 + y2)2 = xy
⇒ x4 + y4 + 2x2y2 = xy
Differentiating both sides w.r.t. x
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.58. If ax2 + 2hxy + by2 + 2gx + 2fy + c = 0, then show thatNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Ans.
Given that: ax2 + 2hxy + by2 + 2gx + 2fy + c = 0.
Differentiating both sides w.r.t. x
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Now, differentiating the given equation w.r.t. y.
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev Hence, proved.

Q.59. IfNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRevprove that
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Ans.
Given that: x = ex/y 
Taking log on both the sides,
log x = log ex/y
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Differentiating both sides w.r.t. x
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRevNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.60. If yx = ey − x , prove thatNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Ans.
Given that: yx = ey – x
Taking log on both sides  log yx = log ey – x
⇒x log y = (y – x) log e
⇒ x log y = y – x [∵ log e = 1]
⇒ x log y + x = y
⇒ x (log y + 1) = y
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Differentiating both sides w.r.t. y
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
We know that
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.61. If NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev , show thatNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Ans.
Given thatNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev 
⇒ y = (cos x)yNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Taking log on both sides log y = y.log (cos x)
Differentiating both sides w.r.t. x
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRevHence, proved.

Q.62. If x sin (a + y) + sin a cos (a + y) = 0, prove thatNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Ans.
Given that:  x sin (a + y) + sin a cos (a + y) = 0
⇒ x sin (a + y) = – sin a cos (a + y)
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Differentiating both sides w.r.t. y
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev Hence proved.

Q.63. IfNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev = a (x – y), prove thatNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Ans.
Given that:NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev = a (x – y)
Put x = sin θ and y = sin ϕ.
∴ θ = sin– 1 x and ϕ = sin– 1 y
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
⇒ cos θ + cos ϕ = a(sin θ – sin ϕ)
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
⇒ sin– 1 x – sin– 1 y = 2 cot– 1 a
Differentiating both sides w.r.t. x
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Hence,NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev

Q.64. If y = tan–1x, findNCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRevin terms of y alone.
Ans.
Given that: y = tan– 1 x ⇒ x = tan y
Differentiating both sides w.r.t. y
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
Again differentiating both sides w.r.t. x
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev
NCERT Exemplar- Continuity and Differentiability (Part-1) Notes | EduRev 

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