NCERT Solutions Exercise 5.3: Continuity & Differentiability | Mathematics (Maths) Class 12 - JEE PDF Download

Find dy/dx in the following:
Q1: 2x + 3y = sin x
Ans: 2x+3y=sinx
Differentiating both sides w.r.t. x, we obtain


Q2: 2x + 3y = sin y
Ans: 2x+3y = siny


Q3: ax + by² = cos y
Ans: ax+by² = cos y
Differentiating w.r.t. x, we have


Q4: xy + y² = tan x + y
Ans: Given, xy+y² = tan x+y
Differentiating both sides w.r.t. x, we get



Q5: x² + xy + y² = 100
Ans: x² + xy + y² = 100
Differentiating both sides w.r.t x we get,


Q6: x³ + x²y + xy² + y³ = 81
Ans: Given equation: x³ + x²y + xy² + y³ = 81
Differentiating the above equation wrt. x



Q7: sin² y + cos xy = κ
Ans: We have, sin²y+cosxy = k
Differentiating both sides with respect to x, we obtain

Using chain rule, we obtain

From (1), (2) and (3), we obtain


Q8: sin² x + cos² y = 1
Ans: We have sin² x + cos² y = 1
Differentiating both sides w.r.t. x, we obtain


Q9: y = sin^–1 (2x/(1+x²))
Ans: y = sin^–1 (2x/(1+x²))
Differentiating above equation w.r.t. x, we have



Q10: y = tan^–1 ((3x-x³)/(1 - 3x²)), -(1/√3) < x < (1/√3)
Ans: 
Put x = tanθ


Q11: y = cos^-1((1-x²)(1+x²)), 0 < x <1
Ans: 
Put x=tanθ
 
= cos⁻¹cos(2θ) = 2θ = 2tan⁻¹x


Q12: y = sin^-1((1-x²)/(1+x²)) 0 < x < 1.
Ans:
 

Q13: y = cos^-1(2x/(1+x²)), -1 < x <1.
Ans:


Q14: y = sin^-1(2x√1-x²), -(1/√2) < x < (1/√2)
Ans: L.H.S = sin⁻¹(2x√1−x²)
Substituting x = sinθ

Now,
= sin⁻¹(2sinθcosθ)
= sin⁻¹(sin2θ)

⇒ sin⁻¹(sin2θ)
= 2θ = 2sin⁻¹x
Hence proved.

Q15: y = sec^-1(1/(2x²-1)), 0 < x < (1/√2).
Ans: 
Put x = cosθ ⇒ θ = cos⁻¹x
Then we have,
y = cos⁻¹(2cos²θ−1) = cos⁻¹(cos2θ) = 2θ = 2cos⁻¹x
 .