Exercise 5.2 Class 12 Maths Continuity and Differentiability

NCERT Solutions Class 12 Maths Chapter 5 Ex 5.2 Detailed | LearnCBSEHub

💡 Key Concept: Chain Rule

If $y = f(g(x))$, then the derivative is found by differentiating the outer function first, then multiplying by the derivative of the inner function.

Formula: $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$

Question 01

Differentiate $y = \sin(x^2 + 5)$

Step 1: Identify Functions

Outer function: $\sin(\dots)$, Inner function: $x^2 + 5$

Step 2: Apply Chain Rule

$$\frac{dy}{dx} = \frac{d}{dx}[\sin(x^2 + 5)]$$ $$= \cos(x^2 + 5) \cdot \frac{d}{dx}(x^2 + 5)$$ $$= \cos(x^2 + 5) \cdot (2x)$$

Answer: $2x \cos(x^2 + 5)$

Question 02

Differentiate $y = \cos(\sin x)$

Step 1: Chain Rule Application

$$\frac{dy}{dx} = \frac{d}{dx}[\cos(\sin x)]$$ $$= -\sin(\sin x) \cdot \frac{d}{dx}(\sin x)$$ $$= -\sin(\sin x) \cdot (\cos x)$$

Answer: $-\cos x \sin(\sin x)$

Question 03

Differentiate $y = \sin(ax + b)$

Step 1: Differentiate

Let $u = ax + b$, then $\frac{du}{dx} = a$.

$$\frac{dy}{dx} = \cos(ax + b) \cdot \frac{d}{dx}(ax + b)$$ $$= \cos(ax + b) \cdot a$$

Answer: $a \cos(ax + b)$

Question 04

Differentiate $y = \sec(\tan x)$

Step 1: Outer Function (Sec)

The derivative of $\sec u$ is $\sec u \tan u$. Here $u = \tan x$.

Step 2: Inner Function (Tan)

$$\frac{dy}{dx} = \sec(\tan x)\tan(\tan x) \cdot \frac{d}{dx}(\tan x)$$ $$= \sec(\tan x)\tan(\tan x) \cdot \sec^2 x$$

Answer: $\sec(\tan x)\tan(\tan x)\sec^2 x$

Question 05

Differentiate $y = \frac{\sin(ax + b)}{\cos(cx + d)}$

Step 1: Apply Quotient Rule

Using $\left(\frac{u}{v}\right)’ = \frac{v u’ – u v’}{v^2}$

Let $u = \sin(ax+b) \implies u’ = a\cos(ax+b)$
Let $v = \cos(cx+d) \implies v’ = -c\sin(cx+d)$

$$\frac{dy}{dx} = \frac{\cos(cx+d) \cdot [a\cos(ax+b)] – \sin(ax+b) \cdot [-c\sin(cx+d)]}{\cos^2(cx+d)}$$

Answer: $\frac{a\cos(ax+b)\cos(cx+d) + c\sin(ax+b)\sin(cx+d)}{\cos^2(cx+d)}$

Question 06

Differentiate $y = \cos(x^3) \cdot \sin^2(x^5)$

Step 1: Apply Product Rule

Using $(uv)’ = u v’ + v u’$

Let $u = \cos(x^3) \implies u’ = -\sin(x^3) \cdot 3x^2$
Let $v = \sin^2(x^5) \implies v’ = 2\sin(x^5)\cos(x^5) \cdot 5x^4$

Step 2: Combine

$$\frac{dy}{dx} = \cos(x^3)[10x^4 \sin(x^5)\cos(x^5)] + \sin^2(x^5)[-3x^2 \sin(x^3)]$$

Answer: $10x^4 \cos(x^3)\sin(x^5)\cos(x^5) – 3x^2 \sin(x^3)\sin^2(x^5)$

Question 07

Differentiate $y = (\cot x)^2$

Step 1: Chain Rule

Treat as $u^2$ where $u = \cot x$.

$$\frac{dy}{dx} = 2(\cot x)^{2-1} \cdot \frac{d}{dx}(\cot x)$$ $$= 2\cot x \cdot (-\csc^2 x)$$

Answer: $-2\cot x \csc^2 x$

Question 08

Differentiate $y = \cos(x^2)$

Step 1: Chain Rule

$$\frac{dy}{dx} = -\sin(x^2) \cdot \frac{d}{dx}(x^2)$$ $$= -\sin(x^2) \cdot (2x)$$

Answer: $-2x \sin(x^2)$

Question 09 • Differentiability

Prove $f(x) = |x – 1|$ is not differentiable at $x = 1$

Step 1: Define Function

$f(x) = 1-x$ for $x < 1$ and $x-1$ for $x \ge 1$.

Step 2: Calculate Derivatives

$$LHD (\text{at } x=1) = \frac{d}{dx}(1-x) = -1$$ $$RHD (\text{at } x=1) = \frac{d}{dx}(x-1) = 1$$

Since $LHD \neq RHD$, the derivative does not exist at this point.

Conclusion: Not Differentiable

Question 10 • Greatest Integer Function

Prove $f(x) = [x], 0 < x < 3$ is not differentiable at $x=1, 2$

Reasoning

The Greatest Integer Function $[x]$ has jump discontinuities at every integer value.

$$\text{At } x=1: \quad \lim_{x \to 1^-} [x] = 0, \quad \lim_{x \to 1^+} [x] = 1$$ $$\text{At } x=2: \quad \lim_{x \to 2^-} [x] = 1, \quad \lim_{x \to 2^+} [x] = 2$$

Since the function is discontinuous at these points, it cannot be differentiable.

Conclusion: Not Differentiable at Integers