Full solutions for Class 12 Maths (NCERT) Exercise 5.4. Practice differentiating functions involving the natural exponential ($e^x$) and logarithmic ($\log x$) functions, mastering the Chain Rule and Quotient Rule in combination with trigonometric and inverse functions.

This exercise focuses on differentiating functions involving exponential functions ($e^x$) and logarithmic functions ($\log x$ or $\ln x$), primarily using the Chain Rule, Quotient Rule, and Product Rule.

Rbse Solutions Class 12 Maths Chapter 5 Exercise 5.4: Differentiation

Differentiate the following w.r.t. $x$

1. $y = \frac{e^x}{\sin x}$

Use the Quotient Rule: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} – u \frac{dv}{dx}}{v^2}$.

Here, $u = e^x$ and $v = \sin x$.

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

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

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

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

Use the Chain Rule: $\frac{d}{dx}(e^u) = e^u \cdot \frac{du}{dx}$.

Here, $u = \sin^{-1} x$.

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

Recall: $\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1 – x^2}}$.

$$\mathbf{\frac{dy}{dx} = \frac{e^{\sin^{-1} x}}{\sqrt{1 – x^2}}}$$

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

Use the Chain Rule.

Here, $u = x^3$.

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

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

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

This requires applying the Chain Rule multiple times: $\frac{d}{dx} \sin(u) = \cos(u) \cdot \frac{du}{dx}$.

Here, $u = \tan^{-1} e^{-x}$.

Differentiate $\sin(u)$:$$\frac{dy}{dx} = \cos(\tan^{-1} e^{-x}) \cdot \frac{d}{dx}(\tan^{-1} e^{-x})$$
Differentiate $\tan^{-1}(v)$, where $v = e^{-x}$:$$\frac{d}{dx}(\tan^{-1} e^{-x}) = \frac{1}{1 + (e^{-x})^2} \cdot \frac{d}{dx}(e^{-x})$$
Differentiate $e^{-x}$:$$\frac{d}{dx}(e^{-x}) = e^{-x} \cdot \frac{d}{dx}(-x) = -e^{-x}$$

Combining the steps:

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

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

5. $y = \log (\cos e^x)$

Use the Chain Rule: $\frac{d}{dx} \log(u) = \frac{1}{u} \cdot \frac{du}{dx}$.

Here, $u = \cos e^x$.

Differentiate $\log(u)$:$$\frac{dy}{dx} = \frac{1}{\cos e^x} \cdot \frac{d}{dx}(\cos e^x)$$
Differentiate $\cos(v)$, where $v = e^x$:$$\frac{d}{dx}(\cos e^x) = -\sin e^x \cdot \frac{d}{dx}(e^x)$$
Differentiate $e^x$:$$\frac{d}{dx}(e^x) = e^x$$

Combining the steps:

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

$$\frac{dy}{dx} = -e^x \frac{\sin e^x}{\cos e^x}$$

$$\mathbf{\frac{dy}{dx} = -e^x \tan(e^x)}$$

6. $y = e^x + e^{x^2} + e^{x^3} + e^{x^4} + e^{x^5}$ (assuming the pattern continues to $e^{x^5}$)

Differentiate each term separately using the Chain Rule:

$$\frac{dy}{dx} = \frac{d}{dx}(e^x) + \frac{d}{dx}(e^{x^2}) + \frac{d}{dx}(e^{x^3}) + \frac{d}{dx}(e^{x^4}) + \frac{d}{dx}(e^{x^5})$$

$$\frac{dy}{dx} = e^x \cdot (1) + e^{x^2} \cdot (2x) + e^{x^3} \cdot (3x^2) + e^{x^4} \cdot (4x^3) + e^{x^5} \cdot (5x^4)$$

$$\mathbf{\frac{dy}{dx} = e^x + 2x e^{x^2} + 3x^2 e^{x^3} + 4x^3 e^{x^4} + 5x^4 e^{x^5}}$$

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

Rewrite the function as $y = (e^{\sqrt{x}})^{1/2} = e^{\frac{1}{2}\sqrt{x}}$.

Use the Chain Rule:

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

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

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

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

$$\frac{dy}{dx} = \frac{e^{\sqrt{x}}}{2\sqrt{e^{\sqrt{x}}}} \cdot \frac{1}{2\sqrt{x}}$$

$$\frac{dy}{dx} = \frac{\sqrt{e^{\sqrt{x}}}}{2} \cdot \frac{1}{2\sqrt{x}}$$

$$\mathbf{\frac{dy}{dx} = \frac{\sqrt{e^{\sqrt{x}}}}{4\sqrt{x}}}$$

(Alternative Method using $y = e^{\frac{1}{2}\sqrt{x}}$ is often simpler)

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

$$\frac{dy}{dx} = e^{\frac{1}{2}\sqrt{x}} \cdot \frac{1}{2} \cdot \frac{1}{2} x^{-1/2}$$

$$\frac{dy}{dx} = \frac{e^{\frac{\sqrt{x}}{2}}}{4\sqrt{x}} = \frac{(e^{\sqrt{x}})^{1/2}}{4\sqrt{x}} = \mathbf{\frac{\sqrt{e^{\sqrt{x}}}}{4\sqrt{x}}}$$