# NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability - Exercise 5.7

### 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 5.7

Find the second order derivative of the given functions.

1. x2 + 3x + 2.

Sol. Let y = x2 + 3x + 2

Differentiating twicely w.r.t. x, we get

$$\frac{dy}{dx} = 2x+3\space\text{and}\space\\\frac{d^{2}y}{dx^{2}} =\frac{d}{dx}(2x+3) = 2$$

2. x20.

Sol. Let y = x20

Differentiating twicely w.r.t. x, we get

$$\frac{dy}{dx} = 20 x^{19}\\\text{and\space}\frac{d^{2}y}{dx^{2}} =\frac{d}{dx}(20x^{19})$$

= 20(19x18) = 380x18

3. x cos x.

Sol. Let y = x cos x

Differentiating twicely w.r.t. x, we get

$$\frac{dy}{dx} = \frac{d}{dx}(x cos x)\\ = x\frac{d}{dx}\text{cos x + cos x}\frac{d}{dx}(x)$$

= x(– sin x) + cos x.1

= – x sin x + cos x

$$\begin{pmatrix}\text{Using product rule}\\\frac{d}{dx}(u.v) =u\frac{dv}{dx} +v\frac{du}{dx}\end{pmatrix}\\\text{and}\space\frac{d^{2}y}{dx^{2}} =\frac{d}{dx}(- sin x + cos x)\\=-\frac{d}{dx}(x sin x) + \frac{d}{dx}(\text{cos x})$$

= – {x cos x + sin x.1} – sin x

= – x cos x – 2 sin x

(Using product rule)

4. log x.

Sol. Let y = log x

Differentiating twicely w.r.t. x, we get

$$\frac{dy}{dx} =\frac{1}{x}\space\text{and}\\\space\frac{d^{2}y}{dx^{2}} =\frac{d}{dx}\bigg(\frac{1}{x}\bigg) =\frac{\normalsize-1}{x^{2}}$$

5. x3 log x.

Sol. Let y = x3 log x. Differentiating twicely w.r.t. x,

we get

$$\frac{dy}{dx} = \frac{d}{dx}(x^{3}\text{log x})\\ = x^{3}\frac{d}{dx}(log \space x) + \text{log x}\frac{d}{dx}(x^{3})\\= x^{3}\bigg(\frac{1}{x}\bigg) + \text{(log x)(3x}^{2})$$

= x2(1 + 3 log x)

(Using product rule)

$$\text{and\space}\frac{d^{2}y}{dx^{2}} = \frac{d}{dx}\lbrack x^{2}(1 + 3 log x)\rbrack\\= x^{2}\bigg(0 +\frac{3}{x}\bigg) +\\(1 + 3 log x)(2x)$$

= 3x + 2x(1 + 3 log x)

= x(5 + 6 log x)

6. ex sin 5x.

Sol. Let y = exsin 5x.

Differentiating twicely w.r.t. x,

we get

$$\frac{dy}{dx} = \frac{d}{dx}(e^{x} \text{sin 5x})\\= e^{x}\frac{d}{dx}(\text{sin 5x}) + sin\space 5x\frac{d}{dx}(e^{x})$$

= ex cos 5x.5 + sin 5x ex

(Using product rule)

= ex (5 cos 5x + sin 5x)

$$\text{and}\space\frac{d^{2}y}{dx^{2}} =\frac{d}{dx}\lbrace e^{x}(5cos5x + sin5x)\rbrace\\= e^{x}\frac{d}{dx}\lbrace5 cos 5x + sin 5x\rbrace + \\(5 cos 5x + sin 5x)\frac{d}{dx}e^{x}$$

= ex{– 5 sin 5x.5 + cos 5x.5} + {5 cos 5x + sin 5x)ex

= ex {10 cos 5x – 24 sin 5x}

= 2ex(5 cos 5x – 12 sin 5x)

7. e6x cos 3x.

Sol. Let y = e6x cos 3x

Differentiating twicely w.r.t. x, we get

$$\frac{dy}{dx} = \frac{d}{dx}\lbrace e^{6x} cos 3x\rbrace\\ = e^{6x}\frac{d}{dx}(cos 3x) + cos 3x\frac{d}{dx}(e)^{6x}$$

= e6x(– sin 3x.3) + (cos 3x) e6x.6

= e6x {– 3 sin 3x + 6 cos 3x}

$$\text{and}\space\frac{d^{2}y}{dx^{2}} =\\\frac{d}{dx}\lbrack e^{6x}(-3 sin \space 3x + 6 cos 3x)\rbrack\\= e^{6x}\frac{d}{dx}(-3 sin 3x + 6 cos\space 3x) + \\(-3 sin \space3x + 6 cos \space3x)\frac{d}{dx}e^{6x}$$

= e6x(– 3 cos 3x.3 – 6 sin 3x.3) + (– 3 sin 3x + 6 cos 3x) e6x.6

= e6x{– 9 cos 3x – 18 sin 3x – 18 sin 3x + 36 cos 3x}

= e6x {27 cos 3x – 36 sin 3x)

= 9e6x {3 cos 3x – 4 sin 3x}

8. tan–1 x.

Sol. Let y = tan–1x.

Differentiating twicely w.r.t. x, we get

$$\frac{dy}{dx} = \frac{d}{dx}(\text{tan}^{\normalsize-1}x) \\=\frac{1}{\text{1 +x}^{2}}\\\text{and}\space\frac{d^{2}y}{dx^{2}} =\frac{d}{dx}\bigg(\frac{1}{1 +x^{2}}\bigg)\\\frac{d}{dx}(1 +x^{2})^{\normalsize-1}\\=-1(1 + x^{2})^{\normalsize-2}\frac{d}{dx}(1 + x^{2})\\= \normalsize-1(1 +x^{2})^{\normalsize-2}(0 +2x) \\=\frac{-2x}{(1 +x^{2})^{2}}$$

9. log (log x).

Sol. Let y = log (log x)

Differentiating twicely w.r.t. x, we get

$$\frac{dy}{dx} =\frac{1}{log x}\frac{d}{dx}(\text{log x}) \\=\frac{1}{\text{log x}}.\frac{1}{x} = \frac{1}{x log x}\\\text{and}\space \frac{d^{2}y}{dx^{2}} =\frac{d}{dx}\bigg(\frac{1}{\text{x log x}}\bigg)\\=\frac{d}{dx}(x\space log x)^{\normalsize-1}\\=-1(xlog x)^{\normalsize-2}\frac{d}{dx}(x log x)\\ =-\frac{1}{(xlog x)^{2}}\\\bigg[x\frac{d}{dx}log x +log x\frac{d}{dx}(x)\bigg]$$

$$=\frac{\normalsize-1}{(x log x)^{2}}\bigg(x.\frac{1}{x} + log x.1\bigg)\\=\frac{-(1 + log\space x)}{(x \space log x)^{2}}$$

10. sin (log x).

Sol. Let y = sin (log x)

Differentiating twicely w.r.t. x, we get

$$\frac{dy}{dx} =\text{cos(log x)}\frac{d}{dx}(log x)\\=\text{cos(log x)}\frac{1}{x} =\frac{\text{cos(log x)}}{x}\\\text{and}\space\frac{d^{2}y}{dx^{2}} =\frac{d}{dx}\bigg(\frac{\text{cos(log x)}}{x}\bigg)\\= \\\frac{x\frac{d}{dx}(\text{cos}(log x)) -\text{cos(log x)}\frac{d}{dx}(x)}{x^{2}}\\=\\\frac{x\begin{Bmatrix}\text{- sin}(log \space x).\frac{1}{x}\end{Bmatrix} -\text{cos(log x)}.1}{x^{2}}$$

(Using quotient rule)

$$= \\-\begin{Bmatrix}\frac{\text{sin(log x) + cos (log x)}}{x^{2}}\end{Bmatrix}$$

11. If y = 5 cos x – 3 sin x,

$$\textbf{prove that}\space\frac{\textbf{d}^{\textbf{2}}\textbf{y}}{\textbf{dx}^{\textbf{2}}}\textbf{+ y = 0}\textbf{.}$$

Sol. Given, y = 5 cos x – 3 sin x

Differentiating twicely w.r.t. x, we get

$$\frac{dy}{dx} =\text{-5 sin x - 3 cos x}\\\text{and}\space\frac{d^{2}y}{dx^{2}} =\frac{d}{dx}(-5 sin x -3 cos x)$$

= – 5 cos x + 3 sin x

= – (5 cos x – 3 sin x)

= – y

$$\Rarr\space\frac{d^{2}y}{dx^{2}} + y =0$$

Hence proved.

12. If y = cos–1x, then find

$$\frac{\textbf{d}^{\textbf{2}}\textbf{y}}{\textbf{dx}^{\textbf{2}}}\space\textbf{in terms of y alone.}$$

Sol. Given, y = cos–1x ⇒ x = cos y

Differentiating w.r.t. y, we get

$$\frac{dx}{dy} =-\text{sin y}\\\Rarr\space\frac{dy}{dx} =-\text{cosec y}\space\text{...(i)}$$

Again, differentiating w.r.t. x, we get

$$\frac{d^{2}y}{dx^{2}} =\frac{d}{dx}(- cosec \space y)\\=(-\text{cosec y cot y})\frac{dy}{dx}$$

= cosec y cot y (– cosec y)

= – cot y.cosec2 y [from Eq. (i)]

13. If y = 3 cos (log x) + 4 sin (log x),

show that x2y2 + xy1 + y = 0.

Sol. Given, y = 3 cos (log x) + 4 sin (log x) ...(i)

Differentiating w.r.t. x, we get

$$\frac{dy}{dx} = y_{1}\\= -3\space\text{sin(log x)}\frac{d}{dx}(log x) +\\\text{4 cos(log x)}\frac{d}{dx}\space\text{log x}\\=-\text{3 sin(log x)}\frac{1}{x} +\\\text{4 cos(log x)}\frac{1}{x}\\\bigg(\because\space\frac{dy}{dx} = y_{1}\bigg)$$

Multiplying by x, we get

xy1 = – 3 sin (log x) + 4 cos (log x) ...(ii)

Again differentiating w.r.t. x, we obtain

xy2 + y1.1 =

$$\text{-3 cos(log x)}\frac{d}{dx}(log x) -\\\text{4 sin(log x)}\frac{d}{dx}\text{log x}\\=\\\text{- 3 cos(log x)}\frac{1}{x} -\text{4 sin(log x)}\frac{1}{x}$$

Multiplying throughout by x, we have

x2y2 + xy1 = – [3 cos (log x) + 4 sin (log x)]

[from Eq. (i)]

$$\Rarr\space x^{2}y_{2} + xy_{1} =-y\\\Rarr\space x^{2}y_{2} +xy_{1} +y = 0$$

14. If y = Aemx + Benx, show that

$$\frac{\textbf{d}^{\textbf{2}}\textbf{y}}{\textbf{dx}^{\textbf{2}}} \textbf{= (m+n)}\frac{\textbf{dy}}{\textbf{dx}} +\\\textbf{mmy = 0.}$$

Sol. Given, y = A emx + B enx ...(i)

Differentiating twicely w.r.t. x, we get

$$\frac{dy}{dx} = Ae^{mx}\frac{d}{dx}(mx) +\\Be^{nx}\frac{d}{dx}(nx)$$

= Aemx + Benx n       ...(ii)

$$\Rarr\space\frac{d^{2}y}{dx^{2}} =m^{2}Ae^{mx} + n^{2} Be^{nx}\\\text{...(iii)}$$

Using equations (i), (ii) and (iii), we have

$$\frac{d^{2}y}{dx^{2}}-(m+n)\frac{dy}{dx} +mny$$

= m2 Aemx + n2 Benx – (m + n) {Amemx + nBenx} + mn {Aemx + Benx}

$$\text{i.e,\space}\frac{d^{2}y}{dx^{2}}-(m+n)\frac{dy}{dx} +mny = 0.$$

Hence proved.

15. If y = 500 e7x + 600 e–7x,

$$\textbf{show that}\space\frac{\textbf{d}^{\textbf{2}}\textbf{y}}{\textbf{dx}^{\textbf{2}}} \textbf{= 49 y.}$$

Sol. Given, y = 500e7x + 600e–7x ...(i)

Differentiating twicely w.r.t. x, we get

$$\frac{dy}{dx} = 500 e^{7x} \frac{d}{dx}(7x) +\\ 600 e^{\normalsize-7x}\frac{d}{dx}(-7x)$$

= 500 e7x.600 e–7x.(– 7)

$$\text{and}\space\frac{d^{2}y}{dx^{2}} =(7×500)e^{7x}.7-\\(7×600) e^{\normalsize-7x}(\normalsize-7)$$

= 49(500 e7x + 600e–7x)

$$\Rarr\space \frac{d^{2}y}{dx^{2}} = 49 y.$$

Hence proved.

16. If ey(x + 1) = 1, show that

$$\frac{\textbf{d}^{\textbf{2}}\textbf{y}}{\textbf{dx}^{\textbf{2}}} =\bigg(\frac{\textbf{dy}}{\textbf{dx}}\bigg)^{\textbf{2}}\textbf{.}$$

Sol. Given, ey (x + 1) = 1 ...(i)

$$\Rarr\space e^{y} =\frac{1}{x+1}$$

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

$$e^{y}\frac{dy}{dx} = -\frac{1}{(x+1)^{2}}\\\Rarr\space \bigg(\frac{1}{x+1}\bigg)\frac{dy}{dx} =-\frac{1}{(x+1)^{2}}\\\bigg(\because\space e^{y} =\frac{1}{x+1}\bigg)\\\Rarr\space\frac{dy}{dx} =-\frac{x+1}{(x+1)^{2}} =\\-\frac{1}{x+1}\space\text{...(ii)}$$

Differentiating again w.r.t. x, we get

$$\frac{d^{2}y}{dx^{2}} =\frac{1}{(x+1)^{2}}\\\text{or}\space\frac{d^{2}y}{dx^{2}} =\bigg(-\frac{1}{x+1}\bigg)^{2} \\=\bigg(\frac{dy}{dx}\bigg)^{2}$$

Hence proved.

17. If y = (tan–1x)2, show that (x2 + 1)2 y2 + 2x(x2 + 1) y1 = 2.

Sol. Given, y = (tan–1 x)2

Differentiating w.r.t. x, we get

$$\frac{dy}{dx} = 2\text{tan}^{\normalsize-1}x\frac{d}{dx}(\text{tan}^{\normalsize-1}x)\\= 2(\text{tan}^{\normalsize-1}x)\frac{1}{1 +x^{2}}$$

or (1 + x2)y1 = 2 tan–1x

$$\bigg(\because\space\frac{dy}{dx} = y_{1}\bigg)$$

Again differentiating w.r.t. x, we get

$$(1 + x^{2})\frac{dy_{1}}{dx} + y_{1}\frac{d}{dx}(1 +x^{2}) \\=\frac{2}{1 +x^{2}}\\\Rarr\space (1 +x^{2})y_{2} + y_{1}(0+2x)\\=\frac{2}{1 +x^{2}}\\\Rarr\space (1 +x^{2})^{2}y_{2} + 2x(1 +x^{2})y_{1}=2\\\bigg(\because\space\frac{d}{dx}(y_{1}) =y_{2}\bigg)$$