# Class 12 Maths NCERT Solutions for Chapter 5 Continuity and Differentiability Exercise 5.2

### Continuity and Differentiability Exercise 5.2 Solutions

**1. Differentiate the functions with respect to x. sin (x^{2} + 5)**

**Solution**

Let f(x) = sin (x^{2} + 5), u(x) = x^{2} + 5, and v(t) = sint

Then, (v o u) (x) = v(u (x)) = v(x^{2} + 5) = sin(x^{2} + 5) = f(x)

Thus, f is a composite of two functions.

Put t = u(x) = x^{2} + 5

Then, we obtain

dv/dt = d/dt (sin t) = cost = cos (x^{2} + 5)

dt/dx = d/dx (x^{2} + 5) = d/dx (x^{2} ) + d/dx (5) = 2x + 0 = 2x

Therefore by chain rule,

df/dx = dv/dt . dt/dx = cos (x^{2} + 5) × 2x =2x cos(x^{2} + 5) **Alternate method:**

**2. Differentiate the functions with respect to x.cos (sin x)**

**Solution**

Let f(x) = cos (sin x), u(x) = sin x, and v(t) = cos t

Then, (vou)(x) = v(u(x)) = v(sin x) = cos (sin x) = f(x)

Thus, f is a composite function of two functions.

Put t = u (x) = sinx

∴ dv/dt = d/dt [cos t] = -sin t = -sin (sin x)

dt/dx = d/dx (sin x) = cos x

By chain rule,

df/dx = dv/dt . dt/dx = -sin (sin x). cos x = -cos x sin(sin x) **Alternate method:**

d/dx [cos (sin x)] = -sin(sin x). d/dx (sin x) = -sin(sin x).cos x = -cos x sin(sin x)

**3. Differentiate the functions with respect t x.sin (ax + b) **

**Solution**

Let f(x) = sin (ax + b), u(x) = ax + b, and v(t) = sin t

Then, (vou)(x) = v (u (x)) = v (ax + b) = sin(ax + b) = f(x)

Thus, f is a composite function of two functions, u and v.

Put t = u(x) = ax + b

Therefore,

dv/dt = d/dt (sin t) = cos t = cos (ax + b)

dt/dx = d/dx (ax + b) = d/dx (ax) + d/dx (b) = a + 0 = a

Hence by chain rule, we obtain

df/dx = dv/dt . dt/dx = cos (ax + b) . a = a cos (ax + b) **Alternate method:**

**4. Differentiate the function with respect to x. sec(tan(√x))**

**Solution**

Let f(x) = sec (tan√x), u(x) = √x, v(t) = tan t, and w(s) = sec s

Then, (wovou) (x) = w[v(u(x))] = w[v(√x)] = w(tan√x) = sec (tan√x) = f(x)

Thus, f is a composite function of three functions, u, v, and w.

Put s = v(t) = tan t and t = u(x) = √x

Then, dw/ds = d/ds (sec s) = sec s tan s = sec (tan t). tan (tan t) **[s = tan t] **

= sec (tan √x) . tan (tan √x) **[t = √x]**

**5. Differentiate the functions with respect to x. sin (ax + b)/cos (cx + d)**

**Solution**

The given function is , where g(x) = sin (ax + b) and

h(x) = cos (cx + d)

∴ f ' = (g'h - gh')/h^{2}

Consider g(x) = sin (ax + b)

Let u(x) = ax + b, v(t) = sin t

Then, (vou) (x) = v(u(x)) = v(ax + b) = sin (ax + b) = g(x)

∴ g is a composite function of two functions, u and v.

Put t = u(x) = ax + b

Consider h(x) = cos (cx + d)

Let p(x) = cx + d, q(y) = cos y

Then, (qop)(x) = q (p(x)) = q(cx + d) = cos (cx + d) = h(x)

h is a composite function of two functions, p and q.

Put y = p(x) = cx +d

**6. Differentiate the functions with respect to x. cos x ^{3} . sin^{2} (x^{3} )**

**Solution**

The given function is cos x^{3} . sin^{2} (x^{3})

d/dx [cos x^{3} .sin^{2} (x^{5})] = sin^{2} (x^{5}) × d/dx( cos x^{3}) + cos x^{3} × d/dx[sin^{2} (x^{5})]

= sin^{2} (x^{5}) × (-sinx^{3}) × d/dx(x^{3}) + cos x^{3} × 2 sin(x^{5}). d/dx [sinx^{5}]

= -sinx^{3} sin^{2} (x^{5}) × 3x^{2} + 2 sinx^{5} cos x^{3} . cosx^{5} × d/dx(x^{5})

= - 3x^{2} sinx^{3} . sin^{2} (x^{5}) + 2 sinx^{5} cos x^{5} cos x^{3} × 5x^{4}

= 10 x^{4} sin x^{5} cos x^{5} cos x^{3} - 3x^{2} sin x^{3} sin^{2} (x^{5})

**7. Differentiate the functions with respect to x. 2√cot(x ^{2} )**

**Solution**

**8. Differentiate the functions with respect to x.**

**cos (√x)**

**Solution**

Also, let u(x) = √x

And, v(t) = cos t

Then, (vou)(x) = v(u(x))

= v(√x)

= cos √x

= f(x)

Clearly, f is a composite function of two functions, u and v, such that

t = u(x) = √x

**9. Prove that the function f given by f(x) = |x - 1|, x ∊ R is not differentiable at x = 1.**

**Solution**

It is known that a function f is differentiable at a point x = c in its domain if both

To check the differentiability of the given function at x = 1,

consider the left 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

**10. Prove that the greatest integer function defined by f(x) = [x], 0 < x < 3 is not differentiable at x = 1 and x = 2.**

**Solution**

It is known that a function f is differentiable at a point x = c in its domain if both

To check the differentiability of the given function at x = 1, consider the left hand limit of f at x = 1

Since the left and right hand limits of f at x = 2 are not equal, f is not differentiable at x = 2