NCERT Solutions Exercise 5.2: Continuity & Differentiability

Differentiate the functions with respect to x in Exercises 1 to 8.
Q1: sin (x² + 5)
Ans: Given y = sin(x²+5)

x²+5=u
du/dx = 2x


Q2: cos (sin x)
Ans: Let y = cos (sin x)
Differentiate both sides w.r.t x, we get
dy/dx = d/dx  (cos (sin x))=-sin(sinx)d/dx (sin x)(By chain rule)
= - sin (sin x) cos x = -cos x sin (sin x)

Q3: sin (ax + b)
Ans: Let y=sin(ax+b)
Thus using chain rule,


Q4: sec (tan(√x))
Ans: Let y = sec (tan√x)
Differentiate both sides w.r.t. x, we get


Q5: sin (ax+b)/cos (cx+d)
Ans: We have, f(x)=
Thus using quotient rule and chain rule simualtaneously,


Q6: cos x³ . sin² (x⁵)
Ans: Let y = cos³ sin² (x⁵)
Differentiate both sides w.r.t. x,

(Using chain rule)
= (cos x³)(2sin x⁵)(cos x⁵)(5x⁴)-sin²(x⁵)(sin x³)(3x²)

Q7: 2√cot(x²)  
Ans:
 

(using sin 2θ=2 sin θ cos θ)

Q8: cos(√x)  
Ans: Compute the derivative:
As we know, the formula for the first principle,

We can write that


Hence the required derivative is .

Q9: Prove that the function f given by f(x) = |x - 1|, x ∈ R is not differentiable at x = 1.
Ans:  The given function is f(x) = |x - 1|, x ∈ R.
It is known that a function f is differentiable at point x=c in its domain if both
 are finite and equal.
To check the differentiability of the function at x=1, 
Consider the left hand limit of f at  x = 1

Consider the right hand limit of f at x - 1

Since the left and right hand limits of f at x=1 are not equal, f is not differentiable at x=1. 

Q10: Prove that the greatest integer function defined by f(x) = [x], 0 < x < 3 is not differentiable at x = 1 and x = 2.
Ans: f(x) = [x], 0<x<3
At x = 1
f(x) is differentiable at x = 1y if
 
∞ ≠ 0
∴ f(x) is not differentaible at x = 1
For x=2

∞ ≠ 0
∴ f(x) is not differentaible at x = 2.