NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability - Exercise 5.4

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Access Exercises of Class 12 Maths Chapter 5 – Continuity and Differentiability

Exercise 5.1 Solutions: 34 Questions (Short Answers)

Exercise 5.2 Solutions: 10 Questions (Short Answers)

Exercise 5.3 Solutions: 15 Questions ( Short Answers)

Exercise 5.4 Solutions: 10 Questions (Short Answers)

Exercise 5.5 Solutions: 18 Questions ( Short Answers)

Exercise 5.6 Solutions: 11 Questions (Short Answers)

Exercise 5.7 Solutions: 17 Questions (Short Answers)

Exercise 5.8 Solutions: 6 Questions (Short Answers)

Miscellaneous Exercise Solutions: 23 Questions (6 Long, 17 Short)

Exercise 5.4

Direction (Q. 1 to 10) : Differentiate the following questions w.r.t. x.

$$\textbf{1.\space}\frac{\textbf{e}^{\textbf{x}}}{\textbf{sin x}}.\\\textbf{Sol.\space}\text{Let y =}\frac{e^{x}}{\text{sin x}}.$$

Differentiate both sides w.r.t. x, we get

$$\frac{dy}{dx} =\frac{d}{dx}\bigg(\frac{e^{x}}{\text{sin x}}\bigg)$$

$$\begin{bmatrix}\text{Using quotient rule,}\\\frac{d}{dx}\bigg(\frac{u}{v}\bigg) =\frac{v\frac{d}{dx}u - u\frac{d}{dx}v}{v^{2}}\end{bmatrix}\\=\frac{\text{sin x}\frac{d}{dx}e^{x}- e^{x}\frac{d}{dx}\text{sin x}}{\text{sin}^{2}x}\\=\frac{\text{sin x e}^{x} - e^{x}\text{cos x}}{\text{sin}^{2}x}\\=\frac{e^{x}(\text{sin x - cos x})}{\text{sin}^{2}x},$$

x ≠ nπ, n∈Z.

$$\textbf{2. e}^{\textbf{sin}^{\normalsize-1}}\textbf{x .}$$

$$\textbf{Sol.\space}\text{Let y = e}^{sin^{\normalsize-1}} x$$

Differentiate both sides w.r.t. x, we get

$$\Rarr\space\frac{dy}{dx} =\frac{d}{dx}(e^{sin^{\normalsize-1}}x)\\= e^{\text{sin}^{\normalsize-1}x}\frac{d}{dx}\text{sin}^{\normalsize-1}x\\\begin{pmatrix}\text{Using chain rule}\\\frac{d}{dx}e^{ax} = e^{ax}\frac{d}{dx}(ax)\end{pmatrix}\\= e^{\text{sin}^{\normalsize-1}x}×\frac{1}{\sqrt{1 - x^{2}}}\\=\frac{e^{sin^{\normalsize-1}x}}{\sqrt{1 - x^{2}}}, x\epsilon(-1,1)\\\begin{bmatrix}\because\space\frac{1}{\sqrt{1 - x^{2}}}\space\text{is defined for}\\x\epsilon (-1,1)\end{bmatrix}$$

$$\textbf{3.\space e}^{\textbf{x}^{\textbf{3}}}\textbf{.}\\\textbf{Sol.\space}\text{Let y = e}^{x^{3}}$$

Differentiate both sides w.r.t. x, we get

$$\Rarr\space\frac{dy}{dx} = \frac{d}{dx}(e^{x^{3}}) = e^{x^{3}}.\frac{d}{dx}(x^{3})\\ = e^{x^{3}}(3x^{2}) = 3x^{2}e^{x^{3}}\\\begin{pmatrix}\text{Using chain rule}\\\frac{d}{dx}e^{ax} = e^{ax}\frac{d}{dx}(ax)\end{pmatrix}$$

4. sin(tan^–1 e^–x).

Sol. Let y = sin(tan^–1e^–x)

Differentiate both sides w.r.t. x, we get

$$\Rarr\space\frac{dy}{dx} = \frac{d}{dx}\lbrack\text{sin}(\text{tan}^{\normalsize-1}(e^{\normalsize-x}))\rbrack\\=\text{cos}\lbrack\text{tan}^{\normalsize-1}(e^{\normalsize-x})\rbrack\\\frac{d}{dx}\lbrack\text{tan}^{\normalsize-1}(e^{\normalsize-x})\rbrack$$

(Using chain rule)

$$= \text{cos}\lbrack\text{tan}^{\normalsize-1}(e^{\normalsize-x})\rbrack\\\frac{1}{1 + (e^{\normalsize-x})^{2}}\frac{d}{dx}(e^{\normalsize-x})\\=\text{cos}\lbrack\text{tan}^{\normalsize-1}(e^{\normalsize-x})\rbrack\\\frac{1}{1 +e^{\normalsize-2x}}(-e^{\normalsize-x})\\=-\frac{e^{\normalsize-x}\text{cos}(tan^{\normalsize-1}e^{\normalsize-x})}{\text{1 +e}^{-2x}}$$

5. log (cos e^x).

Sol. Let y = log (cos e^x).

Differentiate both sides w.r.t. x, we get

$$\frac{dy}{dx} = \frac{d}{dx}\lbrack\text{log}\lbrace\text{cos}(e^{x})\rbrace\rbrack\\=\frac{1}{\text{cos}(e^{x})}\frac{d}{dx}\lbrack\text{cos}(e^{x})\rbrack$$

(Using chain rule)

$$=\frac{1}{\text{cos}(e^{x})}\lbrace-\text{sin}(e^{x})\rbrace\frac{d}{dx}(e^{x})$$

(Using chain rule)

= – tan (e^x).e^x = – e^x tan(e^x)

6. e^x + e^x² + .....+ e^x⁵

Sol. Let y = e^x + e^x² + ..... + e^x⁵

Differentiate both sides w.r.t. x, we get

$$\frac{d}{dx}y = \frac{d}{dx}\\\lbrace e^{x} + e^{x^{2}}+e^{x^{3}} + e^{x^{4}} + e^{x^{5}}\rbrace\\=\space\frac{d}{dx}(e^{x}) +\frac{d}{dx}(e^{x^{2}}) + \frac{d}{dx}(e^{x^{3}})+\\\frac{dv}{dx}(e^{x^{4}}) + \frac{d}{dx}(e^{x^{5}})\\ = e^{x} + e^{x^{2}}\frac{d}{dx}(x^{2}) +e^{x^{3}}\frac{d}{dx}(x^{3})+\\e^{x^{4}}\frac{d}{dx}(x^{4}) + e^{x^{5}}\frac{d}{dx}(x^{5}) $$

(Using chain rule)

= e^x + e^x²(2x)+ e^x³(3x²) + e^x⁴(4x³) + e^x²(5^x⁴)

= e^x + 2xe^x² +3x²e^x³ + 4x³e^x⁴ + 5x⁴e^x⁵

$$\textbf{7.\space}\sqrt{\textbf{e}^{\sqrt{\textbf{x}}},\space} \textbf{x}\gt\textbf{0}\\\textbf{Sol.\space} \text{Let y =}(e^{\sqrt{x}})^{\frac{1}{2}}$$

Differentiate both sides w.r.t.x,

$$\Rarr\space\frac{dy}{dx} =\frac{1}{2}(e\sqrt{x})^{\frac{1}{2}-1}\frac{d}{dx}e^{\sqrt{x}}\\\Rarr\space\frac{dy}{dx} =\frac{1}{2}(e\sqrt{x})^{-\frac{1}{2}}.e\sqrt{x}.\frac{d}{dx}(\sqrt{x})\\\Rarr\space\frac{dy}{dx} = \frac{1}{2}\frac{e\sqrt{x}}{\sqrt{e^{\sqrt{x}}}}×\frac{1}{2\sqrt{x}}\\=\frac{e^{\sqrt{x}}}{4\sqrt{x}\sqrt{e^{\sqrt{x}}}} = \frac{e^{\sqrt{x}}}{4\sqrt{xe^{\sqrt{x}}}}$$

8. log(log x), x > 1.

Sol. Let y = log(log x)

Differentiate both sides w.r.t. x,

$$\frac{dy}{dx} = \frac{d}{dx}(\text{log}(log \space x))\\=\frac{1}{\text{log x}}\begin{Bmatrix}\frac{d}{dx}(\text{log x})\end{Bmatrix}\\\Rarr\space\frac{dy}{dx} = \frac{1}{\text{log x}}\frac{1}{x}\\=\frac{1}{x log x}, x\gt 1$$

$$\textbf{9.\space}\frac{\textbf{cos x}}{\textbf{log x}}, \textbf{x}\gt \textbf{0.}\\\textbf{Sol.\space}\text{Let y =}\frac{\text{cos x}}{\text{log x}}$$

Differentiate both sides w.r.t. x,

$$\Rarr\space\frac{dy}{dx} = \frac{d}{dx}\bigg(\frac{\text{cos x}}{\text{log x}}\bigg),\\x\gt0\\\Rarr\space\frac{dy}{dx} =\\\frac{(\text{log x})\frac{d}{dx}(\text{cos x}) - \text{cos x}\frac{d}{dx}(\text{log x})}{(\text{log x})^{2}}\\\begin{pmatrix}\because\space\frac{d}{dx}\bigg(\frac{u}{v}\bigg) =\frac{v\frac{d}{dx}(u) -u\frac{d}{dx}(v)}{v^{2}}\end{pmatrix}\\\Rarr\space\frac{dy}{dx} =\\\frac{(\text{(log x) (-\text{sin x}) - \text{cos x})}\bigg(\frac{1}{x}\bigg)}{\text{(log x)}^{2}}$$

$$\Rarr\frac{dy}{dx} =-\bigg[\frac{\text{x sin x log x + cos x}}{x(log x)^{2}}\bigg],\\ x\gt 0$$

10. cos (log x + e^x), x > 0.

Sol. Let y = cos (log x + e^x)

Differentiate both sides w.r.t. x, we get

$$\Rarr\space\frac{dy}{dx}= \frac{d}{dx}\lbrace\text{cos}(\text{log x + e}^{x})\rbrace\\\Rarr\space\frac{dy}{dx} =-\text{sin}(\text{log x +e}^{x})\\\frac{d}{dx}(\text{log x + e}^{x})$$

(Using chain rule)

$$\Rarr\space\frac{dy}{dx} =-\text{sin}(\text{log x +e}^{x})\\\bigg(\frac{1}{x} + e^{x}\bigg)\\\Rarr\space\frac{dy}{dx} =\\\frac{-(xe^{x} +1)\text{sin}(\text{log x +e}^{x})}{x}$$

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