Exercise 5.4 Class 12 Maths Continuity and Differentiability

NCERT Solutions Class 12 Maths Chapter 5 Ex 5.4 Detailed | LearnCBSEHub

💡 Key Formulas

1. $\frac{d}{dx}(e^x) = e^x$

2. $\frac{d}{dx}(\log x) = \frac{1}{x}$

3. Chain Rule: $\frac{d}{dx}(e^{g(x)}) = e^{g(x)} \cdot g'(x)$

Question 01

Differentiate $y = \frac{e^x}{\sin x}$

Step 1: Apply Quotient Rule

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

Step 2: Simplify

$$\frac{dy}{dx} = \frac{e^x(\sin x – \cos x)}{\sin^2 x}$$

Answer: $\frac{e^x(\sin x – \cos x)}{\sin^2 x}$

Question 02

Differentiate $y = e^{\sin^{-1} x}$

Step 1: Apply Chain Rule

Outer function: $e^{(\dots)}$, Inner function: $\sin^{-1} x$

$$\frac{dy}{dx} = e^{\sin^{-1} x} \cdot \frac{d}{dx}(\sin^{-1} x)$$ $$\frac{dy}{dx} = e^{\sin^{-1} x} \cdot \frac{1}{\sqrt{1 – x^2}}$$

Answer: $\frac{e^{\sin^{-1} x}}{\sqrt{1 – x^2}}$

Question 03

Differentiate $y = e^{x^3}$

Step 1: Apply Chain Rule

$$\frac{dy}{dx} = e^{x^3} \cdot \frac{d}{dx}(x^3)$$ $$\frac{dy}{dx} = e^{x^3} \cdot (3x^2)$$

Answer: $3x^2 e^{x^3}$

Question 04

Differentiate $y = \sin(\tan^{-1} e^{-x})$

Step 1: Nested Chain Rule

Order: $\sin \to \tan^{-1} \to e^{-x} \to -x$

$$\frac{dy}{dx} = \cos(\tan^{-1} e^{-x}) \cdot \frac{d}{dx}(\tan^{-1} e^{-x})$$ $$= \cos(\tan^{-1} e^{-x}) \cdot \frac{1}{1+(e^{-x})^2} \cdot \frac{d}{dx}(e^{-x})$$ $$= \cos(\tan^{-1} e^{-x}) \cdot \frac{1}{1+e^{-2x}} \cdot (-e^{-x})$$

Answer: $-\frac{e^{-x} \cos(\tan^{-1} e^{-x})}{1+e^{-2x}}$

Question 05

Differentiate $y = \log(\cos e^x)$

Step 1: Chain Rule

$$\frac{dy}{dx} = \frac{1}{\cos e^x} \cdot \frac{d}{dx}(\cos e^x)$$ $$= \frac{1}{\cos e^x} \cdot (-\sin e^x) \cdot \frac{d}{dx}(e^x)$$ $$= -\tan(e^x) \cdot e^x$$

Answer: $-e^x \tan(e^x)$

Question 06

Differentiate $y = e^x + e^{x^2} + \dots + e^{x^5}$

Step 1: Differentiate Term by Term

$$\frac{dy}{dx} = \frac{d}{dx}(e^x) + \frac{d}{dx}(e^{x^2}) + \dots + \frac{d}{dx}(e^{x^5})$$ $$\frac{d}{dx}(e^{x^n}) = e^{x^n} \cdot nx^{n-1}$$

Answer: $e^x + 2x e^{x^2} + 3x^2 e^{x^3} + 4x^3 e^{x^4} + 5x^4 e^{x^5}$

Question 07

Differentiate $y = \sqrt{e^{\sqrt{x}}}, x > 0$

Step 1: Rewrite Function

$y = (e^{\sqrt{x}})^{1/2} = e^{\frac{1}{2}\sqrt{x}}$

Step 2: Differentiate

$$\frac{dy}{dx} = e^{\frac{1}{2}\sqrt{x}} \cdot \frac{d}{dx}\left(\frac{1}{2}\sqrt{x}\right)$$ $$= \sqrt{e^{\sqrt{x}}} \cdot \frac{1}{2} \cdot \frac{1}{2\sqrt{x}}$$

Answer: $\frac{e^{\sqrt{x}}}{4\sqrt{x e^{\sqrt{x}}}} \text{ or } \frac{\sqrt{e^{\sqrt{x}}}}{4\sqrt{x}}$

Question 08

Differentiate $y = \log(\log x), x > 1$

Step 1: Chain Rule

$$\frac{dy}{dx} = \frac{1}{\log x} \cdot \frac{d}{dx}(\log x)$$ $$= \frac{1}{\log x} \cdot \frac{1}{x}$$

Answer: $\frac{1}{x \log x}$

Question 09

Differentiate $y = \frac{\cos x}{\log x}, x > 0$

Step 1: Apply Quotient Rule

$$\frac{dy}{dx} = \frac{\log x \cdot (-\sin x) – \cos x \cdot (\frac{1}{x})}{(\log x)^2}$$ $$\frac{dy}{dx} = \frac{-x \sin x \log x – \cos x}{x(\log x)^2}$$

Answer: $-\frac{x \sin x \log x + \cos x}{x(\log x)^2}$

Question 10

Differentiate $y = \cos(\log x + e^x), x > 0$

Step 1: Chain Rule

$$\frac{dy}{dx} = -\sin(\log x + e^x) \cdot \frac{d}{dx}(\log x + e^x)$$ $$= -\sin(\log x + e^x) \cdot \left(\frac{1}{x} + e^x\right)$$

Answer: $-\left(\frac{1}{x} + e^x\right)\sin(\log x + e^x)$