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# Subjective Type Questions: Differentiation | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

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Q.1. Find the derivative of sin (x2 + 1) with respect to x from first principle.

Ans. 2 x cos (x2 + 1)

Solution. Let f (x) = sin (x2 + 1) then

Q.2. Find the derivative of

Ans. -2/9

Solution.

Q.3.

Ans.

Solution. We have,

(Clearly y is not defined at x = 1)

Q. 4.

Ans.

Solution. We are given

Here y is the sum of two functions and in the second function base as well as power are functions of x. Therefore we will use logarithmic differentiation here.
Let y =  u + v

Differentiating (1) with respect to x, we get

Taking log on both sides on equn (2), we get

log v = x log tan x

Differentiating the above with respect to x, we get

Substituting the value of

Q. 5. Let f be a twice differentiable function such that

f "(x) = – f (x), and f '(x)= g (x), h(x) = [f (x)]2 + [g (x)]2

Find h(10) if h(5) = 11

Ans. 11

Solution. Given that f is twice differentiable such that
f " (x) = – f (x) and f '(x) = g(x)
h (x) = [f (x)]2 + [g (x)]2
To find h (10) when h (5) = 11.
Consider h'(x) = 2f f ' + 2gg' = 2f (x) g(x) + 2g(x)f " (x)
[∴ g(x) = f '(x) ⇒ g' (x) = f ''(x)]
= 2f (x) g (x) + 2g(x) (– f (x))
= 2f (x) g (x) – 2f (x) g (x) = 0

⇒ h is a constant function
∴ h (5) = 11 ⇒ h (10) = 11.

Q. 6. If a be a repeated root of a quadratic equation f(x) = 0 and A(x), B(x) and C(x) be polynomials of degree 3, 4 and 5 respectively, then show that   divisible by f(x), where prime denotes the derivatives.

Solution.

Given that α is a repeated root of quadratic equation f (x) = 0
∴ We must have f (x) = k (x – α)2 ; where k is a non- zero real no.
If we put x = α on both sides of eq. (1); we get

Hence  (x – α) is a factor of F(x) Differentiating eq. (1) w.r. to x, we get

Putting x = α on both sides, we get

[as R1 and R3 are identical]
⇒ (x – α) is a factor of F'(x) al so. Or we can say (x – α)2 is a factor of F(x).
⇒ F (x)  is divisible by f (x).

Q. 7. If x = sec θ – cos θ and y = secn θ - cosn θ , then show that

Solution. We have, x = sec θ – cos θ, y = secn θ – cosn θ

sec θ tan θ + tan θ cos θ = tan θ (sec θ + cos θ)

= n secn θ tan θ +n tan θ cosn θ = n tan θ (secn θ + cosn θ)

...(1)

Also x2 + 4  = (sec θ – cos θ)2 + 4
= sec2 θ + cos2 θ – 2 sec θ cos θ  + 4
= sec2 θ + cos2 θ + 2
= (sec θ + cos θ)2          ...(2)
and y2 + 4 = secn θ – cosn θ)2 + 4
= sec2n θ + cos2n θ – 2 secn θ cosn θ + 4
= sec 2n θ + cos 2n θ + 2
= (secn θ + cosn θ)2     ...(3)

Now we have to prove

[Using (1) and (2)]

= n2 (secn q + cosn q)2
= n2 (y2 + 4)      [From eq. (3)]
= RHS

Q. 8. Find dy/dx at x = – 1, when

Ans. 0

Solution. We have given the function

For x = – 1, we have

....(2)

Differentiating both sides with respect to x, we get

Now differentiating eq. (1), we get

Q. 9.

Solution.

⇒ log y = 3 log x – log (x – a) – log (x – b) – log (x – c)

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