NCERT Solutions for Class 12 Maths Chapter 5 Continuity And Differentiability Ex 5.7

Find the second order derivatives of the functions given in Exercises 1 to 10. 

1. $\text{y=}{\text{x}}^{\text{2}}\text{+3x+2}$.

Ans:  

The given function is $\text{y=}{\text{x}}^{\text{2}}\text{+3x+2}$.

Then, differentiating both sides with respect to $\text{x}$ gives

$\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\text{(}{\text{x}}^{\text{2}}\text{)+}\frac{\text{d}}{\text{dx}}\text{(3x)+}\frac{\text{d}}{\text{dx}}\text{(2)=2x+3+0=2x+3}$

That is,

$\frac{\text{dy}}{\text{dx}}=2\text{x+3}$.

Again, differentiating both sides with respect to $\text{x}$ gives

$\frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{=}\frac{\text{d}}{\text{dx}}\text{(2x+3)=}\frac{\text{d}}{\text{dx}}\text{(2x)+}\frac{\text{d}}{\text{dx}}\text{(3)=2+0=2}$

Hence, $\frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{=2}$.

2. $\mathbf{y}\mathbf{=}{\mathbf{x}}^{\mathbf{20}}$.

Ans: 

The given function is $\text{y=}{\text{x}}^{\text{20}}$.

Then, differentiating both sides with respect to $\text{x}$ gives

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

Again, differentiating both sides with respect to $\text{x}$ gives

$$\frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{=}\frac{\text{d}}{\text{dx}}\text{(20}{\text{x}}^{\text{19}}\text{)=20}\frac{\text{d}}{\text{dx}}\text{(}{\text{x}}^{\text{19}}\text{)=20}\left(19\right){\text{x}}^{\text{18}}\text{=380}{\text{x}}^{\text{18}}$$.

Hence, $$\frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{=380}{\text{x}}^{\text{18}}$$.

3. $\mathbf{y}\mathbf{=}\mathbf{x}\cdot \mathbf{cosx}$.

Ans: 

The given function is $\text{y=x}\text{.cosx}$.

Then, differentiating both sides with respect to $\text{x}$ gives

$\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\text{(x}\text{.cosx)=cosx}\text{.}\frac{\text{d}}{\text{dx}}\text{(x)+x}\frac{\text{d}}{\text{dx}}\text{(cosx)=cosx}\text{.1+x(-sinx)=cosx-xsinx}$

That is, $\frac{\text{dy}}{\text{dx}}\text{=cosx-xsinx}$.

Again, differentiating both sides with respect to $\text{x}$ gives

$\begin{array}{rl}& \frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{=}\frac{\text{d}}{\text{dx}}\text{(cosx-xsinx)=}\frac{\text{d}}{\text{dx}}\text{(cosx)-}\frac{\text{d}}{\text{dx}}\text{(xsinx)}\\ & \text{=-sinx- }[\text{ sinx}\cdot \frac{\text{d}}{\text{dx}}\text{(x)+x}\cdot \frac{\text{d}}{\text{dx}}\text{(sinx) }]\\ & \text{=-sinx-(sinx+xcosx)}\end{array}$

Hence, $\frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{=-(xcosx+2sinx)}$.

4. $\mathbf{y}\mathbf{=}\mathbf{logx}$.

Ans: 

The given function is $\text{y=logx}$.

Then, differentiating both sides with respect to $\text{x}$ gives

$$\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\text{(logx)=}\frac{\text{1}}{\text{x}}$$

Again, differentiating both sides with respect to $\text{x}$ gives

$$\frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{=}\frac{\text{d}}{\text{dx}}\left(\frac{\text{1}}{\text{x}}\right)\text{=}\frac{\text{-1}}{{\text{x}}^{\text{2}}}$$

Hence, $$\frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{=-}\frac{\text{1}}{{\text{x}}^{\text{2}}}$$.

5. $\text{y=}{\text{x}}^{\text{3}}\text{logx}$.

Ans: 

The given function is $\text{y=}{\text{x}}^{\text{3}}\text{logx}$.

Then, differentiating both sides with respect to $\text{x}$ gives

$\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\left[{\text{x}}^{\text{3}}\text{logx}\right]\text{=logx}\text{.}\frac{\text{d}}{\text{dx}}\text{(}{\text{x}}^{\text{3}}\text{)+}{\text{x}}^{\text{3}}\frac{\text{d}}{\text{dx}}\text{(logx)=logx}\text{.3}{\text{x}}^{\text{2}}\text{+}{\text{x}}^{\text{3}}\text{.}\frac{\text{1}}{\text{x}}\text{=logx}\text{.3}{\text{x}}^{\text{2}}\text{+}{\text{x}}^{\text{2}}$

That is, $\frac{\text{dy}}{\text{dx}}\text{=}{\text{x}}^{\text{2}}\text{(1+3logx)}$. 

Again, differentiating both sides with respect to $\text{x}$ gives

$\begin{array}{rl}& \frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{=}\frac{\text{d}}{\text{dx}}\text{(}{\text{x}}^{\text{2}}\text{(1+3logx))}\\ & \text{=(1+3logx)}\text{.}\frac{\text{d}}{\text{dx}}\text{(}{\text{x}}^{\text{2}}\text{)+}{\text{x}}^{\text{2}}\frac{\text{d}}{\text{dx}}\text{(1+3logx)}\\ & \text{=(1+3logx)}\text{.2x+}{\text{x}}^{\text{3}}\text{.}\frac{\text{3}}{\text{x}}\\ & \text{=2x+6logx+3x}\\ & \text{=5x+6xlogx}\end{array}$

Hence, $\frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{=x(5+6logx)}$.

6. $\mathbf{y}\mathbf{=}{\mathbf{e}}^{\mathbf{x}}\mathbf{sin}\mathbf{5}\mathbf{x}$

Ans: 

The given function is $\text{y=}{\text{e}}^{\text{x}}\text{sin5x}$.

Then, differentiating both sides with respect to $\text{x}$ gives

$$\begin{array}{rl}& \frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\left[{\text{e}}^{\text{x}}\text{sin5x}\right]\text{=sinx}\frac{\text{d}}{\text{dx}}\text{(}{\text{e}}^{\text{x}}\text{)+}{\text{e}}^{\text{x}}\frac{\text{d}}{\text{dx}}\text{(sin5x)}\\ & \Rightarrow \frac{\text{dy}}{\text{dx}}\text{=sin5x}\text{.}{\text{e}}^{\text{x}}\text{+}{\text{e}}^{\text{x}}\text{.cos5x}\text{.}\frac{\text{d}}{\text{dx}}\text{(5x)}\end{array}$$

That is, $\frac{\text{dy}}{\text{dx}}\text{=}{\text{e}}^{\text{x}}\text{(sin5x+5cos5x)}$.

Again, differentiating both sides with respect to $\text{x}$ gives

$$\begin{array}{rl}& \frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{=}\frac{\text{d}}{\text{dx}}\left[{\text{e}}^{\text{x}}\text{(sin5x+5cos5x)}\right]\\ & \text{=(sin5x+5cos5x)}\text{.}\frac{\text{d}}{\text{dx}}\text{(}{\text{e}}^{\text{x}}\text{)+}{\text{e}}^{\text{x}}\text{.}\frac{\text{d}}{\text{dx}}\text{(sin5x+5cos5x)}\\ & \text{=(sin5x+5cos5x)(}{\text{e}}^{\text{x}}\text{)+}{\text{e}}^{\text{x}}\left[\text{cos5x}\text{.}\frac{\text{d}}{\text{dx}}\text{(5x)+5(-sin5x)}\text{.}\frac{\text{d}}{\text{dx}}\text{(5x)}\right]\\ & \text{=}{\text{e}}^{\text{x}}\text{(sin5x+5cos5x)+}{\text{e}}^{\text{x}}\text{(5cos5x-25sin5x)}\end{array}$$

 $\text{=}{\text{e}}^{\text{x}}\text{(10cos5x-24sin5x)}$.

Hence, $\frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{=2}{\text{e}}^{\text{x}}\text{(5cox5x-12sin5x)}$.

7. $\text{y=}{\text{e}}^{\text{6x}}\text{cos3x}$.

Ans:  

The given function is $\text{y=}{\text{e}}^{\text{6x}}\text{cos3x}$.

Then, differentiating both sides with respect to $\text{x}$ gives

$\begin{array}{rl}& \frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\text{(}{\text{e}}^{\text{6x}}\text{cos3x)=cos3x }\times \frac{\text{d}}{\text{dx}}\text{(}{\text{e}}^{\text{6x}}\text{)+}{\text{e}}^{\text{6x}}\times \frac{\text{d}}{\text{dx}}\text{(cos3x)}\\ & \Rightarrow \frac{\text{dy}}{\text{dx}}=\text{cos3x }\times {\text{e}}^{\text{6x}}\times \frac{\text{d}}{\text{dx}}\text{(6x)+}{\text{e}}^{\text{6x}}\times \text{ (-sin3x) }\times \frac{\text{d}}{\text{dx}}\text{(3x)}\end{array}$

Therefore,

$\frac{\text{dy}}{\text{dx}}\text{=6}{\text{e}}^{\text{6x}}\text{cos3x-3}{\text{e}}^{\text{6x}}\text{sin3x}$ …… (1)

Again, differentiating both sides with respect to $\text{x}$ gives

$\frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{=}\frac{\text{d}}{\text{dx}}\text{(6}{\text{e}}^{\text{6x}}\text{cos3x-3}{\text{e}}^{\text{6x}}\text{sin3x)=6 }\times \frac{\text{d}}{\text{dx}}\text{(}{\text{e}}^{\text{6x}}\text{cos3x)-3 }\times \frac{\text{d}}{\text{dx}}\text{(}{\text{e}}^{\text{6x}}\text{sin3x)}$

$\text{=6 }\times \left[\text{6}{\text{e}}^{\text{6x}}\text{cos3x-3}{\text{e}}^{\text{6x}}\text{sin3x}\right]\text{-3 }\times [\text{sin3x }\times \frac{\text{d}}{\text{dx}}\text{(}{\text{e}}^{\text{6x}}\text{)+}{\text{e}}^{\text{6x}}\times \frac{\text{d}}{\text{dx}}\text{(sin3x)}]$ [using (1)]

$$\begin{array}{rl}& \text{=36}{\text{e}}^{\text{6x}}\text{cos3x-18}{\text{e}}^{\text{6x}}\text{sin3x-3}[\text{sin3x }\times {\text{e}}^{\text{6x}}\times \text{ 6+}{\text{e}}^{\text{6x}}\times \text{ cos3x-3}]\\ & \text{=36}{\text{e}}^{\text{6x}}\text{cos3x-18}{\text{e}}^{\text{6x}}\text{sin3x-18}{\text{e}}^{\text{6x}}\text{sin3x-9}{\text{e}}^{\text{6x}}\text{cos3x}\end{array}$$

Hence, $$\frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{=9}{\text{e}}^{\text{6x}}\text{(3cos3x-4sin3x)}$$.

8. $\mathbf{y}\mathbf{=}\mathbf{ta}{\mathbf{n}}^{\mathbf{-}\mathbf{1}}\mathbf{x}$.

Ans: 

The given function is $\text{y=ta}{\text{n}}^{\text{-1}}\text{x}$.

Then, differentiating both sides with respect to $\text{x}$ gives

$\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\text{ta}{\text{n}}^{\text{-1}}\text{x=}\frac{\text{1}}{\text{1-}{\text{x}}^{\text{2}}}$

Again, differentiating both sides with respect to $\text{x}$ gives

$\begin{array}{rl}& \frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{=}\frac{\text{d}}{\text{dx}}\left(\frac{\text{1}}{\text{1+}{\text{x}}^{\text{2}}}\right)\text{=}\frac{\text{d}}{\text{dx}}{\text{(1+}{\text{x}}^{\text{2}}\text{)}}^{\text{-1}}\text{=(-1) }\times \text{ (1+}{\text{x}}^{\text{2}}{\text{)}}^{\mathbf{-}\mathbf{2}}\times \frac{\text{d}}{\text{dx}}\text{(1+}{\text{x}}^{\text{2}}\text{)}\\ & \text{=-}\frac{\text{1}}{{\text{(1+}{\text{x}}^{\text{2}}\text{)}}^2}\times \text{ 2x}\end{array}$

Hence, $\frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{=}\frac{\text{-2x}}{{\text{(1+}{\text{x}}^{\text{2}}\text{)}}^2}$.

9. $\text{y=log(logx)}$.

Ans: 

The given function is $\text{y=log(logx)}$.

Now, differentiating both sides with respect to $\text{x}$ gives

$\begin{array}{rl}& \frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}[\text{ log(logx) }]\\ & =\frac{\text{1}}{\text{logx}}\times \frac{\text{d}}{\text{dx}}\text{(logx)}\\ & \Rightarrow \frac{\text{dy}}{\text{dx}}\text{=(xlogx}{\text{)}}^{\text{-1}}\end{array}$

Again, differentiating both sides with respect to $\text{x}$ gives

$\begin{array}{rl}& \frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{=}\frac{\text{d}}{\text{dx}}\left[{\text{(xlogx)}}^{\text{-1}}\right]\text{=(-1) }\times \text{ (xlogx}{\text{)}}^{\text{-2}}\frac{\text{d}}{\text{dx}}\text{(xlogx)}\\ & =\frac{\text{-1}}{{\text{(xlogx)}}^{\text{2}}}\times [\text{logx }\times \frac{\text{d}}{\text{dx}}\text{(x)+x }\times \frac{\text{d}}{\text{dx}}\text{(logx)}]\\ & =\frac{\text{-1}}{{\text{(xlogx)}}^{\text{2}}}\times [\text{logx }\times \text{ 1+x }\times \frac{\text{1}}{\text{x}}]\end{array}$

Hence, $\frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{=}\frac{\text{-(1+logx)}}{{\text{(xlogx)}}^{\text{2}}}$.

10. $\text{y=sin(logx)}$.

Ans: 

 The given function is $\text{y=sin(logx)}$.

Now, differentiating both sides with respect to $\text{x}$ gives

$$\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}[\text{ sin(logx) }]\text{ =cos(logx) }\times \frac{\text{d}}{\text{dx}}\text{(logx)}=\frac{\text{cos(logx)}}{\text{x}}$$

Again, differentiating both sides with respect to $\text{x}$ gives

$$\begin{array}{rl}& \frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{=}\frac{\text{d}}{\text{dx}}\left[\frac{\text{cos(logx)}}{\text{x}}\right]\\ & =\frac{\text{x}\left[\text{cos(logx)}\right]\text{-cos(logx) }\times \frac{\text{d}}{\text{dx}}\text{(x)}}{{\text{x}}^{\text{2}}}\\ & =\frac{\text{x}[\text{-sin(logx) }\times \frac{\text{d}}{\text{dx}}\text{(logx)}]\text{-cos(logx) }\times \text{ 1}}{{\text{x}}^{\text{2}}}\\ & =\frac{\text{-xsin(logx) }\times \frac{\text{1}}{\text{x}}\text{-cos(logx)}}{{\text{x}}^{\text{2}}}\end{array}$$

Hence, $$\frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{=}\frac{\left[\text{-sin(logx)-cos(logx)}\right]}{{\text{x}}^{\text{2}}}$$.

11. If $\text{y=5cosx-3sinx}$, prove that $\frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{+y=0}$. 

Ans: 

The given equation is $\text{y=5cosx-3sinx}$.

Then, differentiating both sides with respect to $\text{x}$ gives

$\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\text{(5cosx)-}\frac{\text{d}}{\text{dx}}\text{(3sinx)=5}\frac{\text{d}}{\text{dx}}\text{(cosx)-3}\frac{\text{d}}{\text{dx}}\text{(sinx)=5(-sinx)-3cosx}$

Therefore, $\frac{\text{dy}}{\text{dx}}\text{=-(5sinx+3cosx)}$.

Again, differentiating both sides with respect to $\text{x}$ gives

$\frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{=}\frac{\text{d}}{\text{dx}}[\text{ -(5sinx+3cosx) }]$

$\begin{array}{rl}& \text{=-}[\text{5 }\times \frac{\text{d}}{\text{dx}}\text{(sinx)+3 }\times \frac{\text{d}}{\text{dx}}\text{(cosx)}]\\ & \text{= }[\text{ 5cosx+3(-sinx) }]\\ & \text{=-y}\end{array}$

That is, $\frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{=-y}$.

Hence, $\frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{+y=0}$.

12. If $\mathbf{y}\mathbf{=}\mathbf{co}{\mathbf{s}}^{\mathbf{-}\mathbf{1}}\mathbf{x}$, Find $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{\mathbf{d}{\mathbf{x}}^{\mathbf{2}}}$ in the terms of $\mathbf{y}$ alone.

Ans: 

The given function is $\text{y=co}{\text{s}}^{\text{-1}}\text{x}$.

Now, differentiating both sides with respect to $\text{x}$ gives

$\frac{\text{dy}}{\text{dx}}\text{=}\frac{\text{d}}{\text{dx}}\text{(co}{\text{s}}^{\text{-1}}\text{x)=}\frac{\text{-1}}{\sqrt{\text{1-}{\text{x}}^{\text{2}}}}\text{=-(1-}{\text{x}}^{\text{2}}{\text{)}}^{\frac{\text{-1}}{\text{2}}}$

Again, differentiating both sides with respect to $\text{x}$ gives

$\begin{array}{rl}& \frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{=}\frac{\text{d}}{\text{dx}}\left[\text{-(1-}{\text{x}}^{\text{2}}{\text{)}}^{\frac{\text{-1}}{\text{2}}}\right]\\ & =\left(\frac{\text{-1}}{\text{2}}\right)\times \text{ (1-}{\text{x}}^{\text{2}}{\text{)}}^{\frac{\text{-3}}{\text{2}}}\times \frac{\text{d}}{\text{dx}}\text{(1-}{\text{x}}^{\text{2}}\text{)}\\ & =\frac{\text{1}}{\sqrt[\text{2}]{{\text{(1-}{\text{x}}^{\text{2}}\text{)}}^{\text{3}}}}\times \text{ (-2x)}\end{array}$

$\Rightarrow \frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{=}\frac{\text{-x}}{\sqrt{{\text{(1-}{\text{x}}^{\text{2}}\text{)}}^{\text{3}}}}$                                                                       …… (1)

Now, $\text{y=co}{\text{s}}^{\text{-1}}\text{x}\Rightarrow \text{x=cosy}$.

Therefore, substituting $\text{x=cosy}$ into equation (1), gives

$\begin{array}{rl}& \frac{{\text{d}}^{\text{2}}\text{x}}{\text{d}{\text{y}}^{\text{2}}}\text{=}\frac{\text{-cosy}}{\sqrt{{\text{(1-co}{\text{s}}^{\text{2}}\text{y)}}^{\text{3}}}}\\ & =\frac{\text{-cosy}}{\text{si}{\text{n}}^{\text{3}}\text{y}}\\ & =\frac{\text{-cosy}}{\text{siny}}\times \frac{\text{1}}{\text{si}{\text{n}}^{\text{2}}\text{y}}\end{array}$

Hence, $\frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{= -coty }\times \text{ cose}{\text{c}}^{\text{2}}\text{y}$.

13. If $\mathbf{y}\mathbf{=}\mathbf{3}\mathbf{cos}\mathbf{(}\mathbf{logx}\mathbf{)}\mathbf{+}\mathbf{4}\mathbf{sin}\mathbf{(}\mathbf{logx}\mathbf{)}$, show that ${\mathbf{x}}^{\mathbf{2}}{\mathbf{y}}_{\mathbf{2}}\mathbf{+}\mathbf{x}{\mathbf{y}}_{\mathbf{1}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}$.

Ans: 

The given equations are $\text{y=3cos(logx)+4sin(logx)}$                               …… (1)

and ${\text{x}}^{\text{2}}{\text{y}}_{\text{2}}\text{+x}{\text{y}}_{\text{1}}\text{+y=0}$                                                                                …… (2)

Then, differentiating both sides of the equation (1) with respect to $\text{x}$ gives

$$\begin{array}{rl}& {\text{y}}_{\text{1}}\text{=3 }\times \frac{\text{d}}{\text{dx}}[\text{ cos(logx) }]\text{ +4 }\times \frac{\text{d}}{\text{dx}}[\text{ sin(logx) }]\\ & \text{=3 }\times [\text{-sin(logx) }\times \frac{\text{d}}{\text{dx}}\text{(logx)}]\text{+4 }\times [\text{cos(logx) }\times \frac{\text{d}}{\text{dx}}\text{(logx)}]\end{array}$$

$${\text{y}}_{\text{1}}\text{=}\frac{\text{-3sin(logx)}}{\text{x}}\text{+}\frac{\text{4cos(logx)}}{\text{x}}\text{=}\frac{\text{4cos(logx)-3sin(logx)}}{\text{x}}$$

Again, differentiating both sides with respect to $\text{x}$ gives

$${\text{y}}_{\text{2}}\text{=}\frac{\text{d}}{\text{dx}}\left(\frac{\text{4cos(logx)-3sin(logx)}}{\text{x}}\right)$$

$$\begin{array}{rl}& \text{=}\frac{\text{x}\cdot \frac{d}{dx}[\text{4 }\{\text{ cos(logx) }\}\text{ - }\{\text{ 3sin(logx) }\}]\text{- }\{\text{ 4cos(logx)-3sin(logx) }\}\times \text{1}}{{\text{x}}^{\text{2}}}\\ & \text{=}\frac{\text{x}\left[\text{-4sin(logx)}\frac{d}{dx}\text{(logx)-3cos(logx)}\frac{d}{dx}\text{(logx)}\right]\text{-4cos(logx)+3sin(logx)}}{{\text{x}}^{\text{2}}}\end{array}$$

$={\displaystyle \frac{\text{x}[\text{-4sin(logx)}\cdot {\displaystyle \frac{\text{1}}{\text{x}}}\text{-3cos(logx)}\cdot {\displaystyle \frac{\text{1}}{\text{x}}}]\text{-4cos(logx)+3sin(logx)}}{{\text{x}}^{\text{2}}}}$

$={\displaystyle \frac{\text{-4sin(logx)-3cos(logx)-4cos(logx)+3sin(logx)}}{{\text{x}}^{\text{2}}}}$

Therefore, ${\text{y}}_{\text{2}}\text{=}{\displaystyle \frac{\text{-sin(logx)-7cos(logx)}}{{\text{x}}^{\text{2}}}}$.

Now, substituting the derivatives ${\text{y}}_1$ ,${\text{y}}_2$ and $\text{y}$ into the LHS of the equation (2) gives

$\begin{array}{rl}& {\text{x}}^{\text{2}}{\text{y}}_{\text{2}}\text{+x}{\text{y}}_{\text{1}}\text{+y}\\ & \text{=}{\text{x}}^{\text{2}}\left(\frac{\text{-sin(logx)-7cos(logx)}}{{\text{x}}^{\text{2}}}\right)\text{+x}\left(\frac{\text{4cos(logx)-3sin(logx)}}{{\text{x}}^{\text{2}}}\right)\text{+3cos(logx)+4sin(logx)}\\ & \text{=-sin(logx)-7cos(logx)+4cos(logx)-3sin(logx)+4sin(logx)}\\ & \text{=0}\end{array}$

Hence, it has been proved that ${\text{x}}^{\text{2}}{\text{y}}_{\text{2}}\text{+x}{\text{y}}_{\text{1}}\text{+y}=0$.

14. If $\mathbf{y}\mathbf{=}\mathbf{A}{\mathbf{e}}^{\mathbf{mx}}\mathbf{+}\mathbf{B}{\mathbf{e}}^{\mathbf{nx}}$, show that $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{\mathbf{d}{\mathbf{x}}^{\mathbf{2}}}\mathbf{-}\mathbf{(}\mathbf{m}\mathbf{+}\mathbf{n}\mathbf{)}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}\mathbf{mny}\mathbf{=}\mathbf{0}$.

Ans: 

The given equations are $\text{y=A}{\text{e}}^{\text{mx}}\text{+B}{\text{e}}^{\text{nx}}$                                     …… (1)

and $$\frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{-(m+n)}\frac{\text{dy}}{\text{dx}}\text{+mny=0}$$                                                      ……. (2)

Then, differentiating both sides of the equation (1) with respect to $\text{x}$ gives

$\frac{\text{dy}}{\text{dx}}\text{=A}\text{.}\frac{\text{d}}{\text{dx}}\text{(}{\text{e}}^{\text{mx}}\text{)+B}\text{.}\frac{\text{d}}{\text{dx}}\text{(}{\text{e}}^{\text{mx}}\text{)=A}\text{.}{\text{e}}^{\text{mx}}\text{.}\frac{\text{d}}{\text{dx}}\text{(mx)+B}\text{.}{\text{e}}^{\text{nx}}\text{.}\frac{\text{d}}{\text{dx}}\text{(nx)=Am}{\text{e}}^{\text{mx}}\text{+Bn}{\text{e}}^{\text{nx}}$

Again, differentiating both sides with respect to $\text{x}$ gives

$\begin{array}{rl}& \frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{=}\frac{\text{d}}{\text{dx}}\text{(Am}{\text{e}}^{\text{mx}}\text{+Bn}{\text{e}}^{\text{nx}}\text{)=Am}\text{.}\frac{\text{d}}{\text{dx}}\text{(}{\text{e}}^{\text{mx}}\text{)+Bn}\text{.}\frac{\text{d}}{\text{dx}}\text{(}{\text{e}}^{\text{nx}}\text{)}\\ & \text{=Am}\text{.}{\text{e}}^{\text{mx}}\text{.}\frac{\text{d}}{\text{dx}}\text{(mx)+Bn}\text{.}{\text{e}}^{\text{nx}}\text{.}\frac{\text{d}}{\text{dx}}\text{(nx)}\end{array}$

Therefore, $\frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{=A}{\text{m}}^{\text{2}}{\text{e}}^{\text{mx}}\text{+B}{\text{n}}^{\text{2}}{\text{e}}^{\text{nx}}$.

Thus, substituting the derivatives ${\text{y}}_1$ ,${\text{y}}_2$ and $\text{y}$ into the LHS of the equation (2) gives

$\begin{array}{rl}& \frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{-(m+n)}\frac{\text{dy}}{\text{dx}}\text{+mny}\\ & \text{=A}{\text{m}}^{\text{2}}\text{e}{\text{x}}^{\text{mx}}\text{+B}{\text{n}}^{\text{2}}{\text{e}}^{\text{nx}}\text{-(m+n)}\text{.(Am}{\text{e}}^{\text{mx}}\text{+Bn}{\text{e}}^{\text{nx}}\text{)+mn(A}{\text{e}}^{\text{mx}}\text{+B}{\text{e}}^{\text{nx}}\text{)}\\ & \text{=A}{\text{m}}^{\text{2}}\text{e}{\text{x}}^{\text{mx}}\text{+B}{\text{n}}^{\text{2}}{\text{e}}^{\text{nx}}\text{-Ame}{\text{x}}^{\text{mx}}\text{+Bmn}{\text{e}}^{\text{nx}}\text{+Amn}{\text{e}}^{\text{mx}}\text{+B}{\text{n}}^{\text{2}}{\text{e}}^{\text{nx}}\text{+Amn}{\text{e}}^{\text{mx}}\text{+Bmn}{\text{e}}^{\text{nx}}\\ & \text{=0}\end{array}$

Thus, it has been proved that $$\frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{-(m+n)}\frac{\text{dy}}{\text{dx}}\text{+mny=0}$$.

15. If $\mathbf{y}\mathbf{=}\mathbf{500}{\mathbf{e}}^{\mathbf{7}\mathbf{x}}\mathbf{+}\mathbf{600}{\mathbf{e}}^{\mathbf{-}\mathbf{7}\mathbf{x}}$, show that $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{\mathbf{d}{\mathbf{x}}^{\mathbf{2}}}\mathbf{=}\mathbf{49}\mathbf{y}$.

Ans: 

The given equation is $\text{y=500}{\text{e}}^{\text{7x}}\text{+600}{\text{e}}^{\text{-7x}}$.                                     …… (1)

Then, differentiating both sides with respect to $\text{x}$ gives

$\begin{array}{rl}& \frac{\text{dy}}{\text{dx}}\text{=500 }\times \text{ (}{\text{e}}^{\text{7x}}\text{)+600 }\times \frac{\text{d}}{\text{dx}}\text{(-7x)}\\ & \text{=500 }\times {\text{e}}^{\text{7x}}\times \frac{\text{d}}{\text{dx}}\text{(7x)+600 }\times {\text{e}}^{\text{-7x}}\times \frac{\text{d}}{\text{dx}}\text{(-7x)}\\ & \text{=3500}{\text{e}}^{\text{7x}}\text{-4200}{\text{e}}^{\text{-7x}}\end{array}$

Again, differentiating both sides with respect to $\text{x}$ gives

$\begin{array}{rl}& \frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{=3500 }\times \frac{\text{d}}{\text{dx}}\text{(}{\text{e}}^{\text{7x}}\text{)-4200 }\times \frac{\text{d}}{\text{dx}}\text{(}{\text{e}}^{\text{-7x}}\text{)}\\ & \text{=3500 }\times {\text{e}}^{\text{7x}}\times \frac{\text{d}}{\text{dx}}\text{(7x)-4200 }\times {\text{e}}^{\text{-7x}}\times \frac{\text{d}}{\text{dx}}\text{(-7x)}\\ & \text{=7 }\times \text{ 3500 }\times {\text{e}}^{\text{7x}}\text{+7 }\times \text{ 4200 }\times {\text{e}}^{\text{-7x}}\\ & \text{=49 }\times \text{ 500}{\text{e}}^{\text{7x}}\text{+49 }\times \text{ 600}{\text{e}}^{\text{-7x}}\\ & \text{=49(500}{\text{e}}^{\text{7x}}\text{+600}{\text{e}}^{\text{-7x}}\text{)}\end{array}$

$\text{=49y}$, using the equation (1).

Thus, it has been proved that $\frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{=49y}$.

16. If ${\mathbf{e}}^{\mathbf{y}}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{=}\mathbf{1}$, show that $\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{\mathbf{d}{\mathbf{x}}^{\mathbf{2}}}\mathbf{=}{\left(\frac{\mathbf{dy}}{\mathbf{dx}}\right)}^{\mathbf{2}}$.

Ans: 

The given equation is ${\text{e}}^{\text{y}}\text{(x+1)=1}$.

Now, ${\text{e}}^{\text{y}}\text{(x+1)=1}\Rightarrow {\text{e}}^{\text{y}}\text{=}\frac{\text{1}}{\text{x+1}}$.

So, taking logarithm bth sides of the equation gives

$\text{y=log}\frac{\text{1}}{\text{(x+1)}}$

Therefore, differentiating both sides with respect to $\text{x}$ gives

$\frac{\text{dy}}{\text{dx}}\text{=(x+1)}\frac{\text{d}}{\text{dx}}\left(\frac{\text{1}}{\text{x+1}}\right)\text{=(x+1) }\times \frac{\text{-1}}{{\text{(x+1)}}^{\text{2}}}\text{=}\frac{\text{-1}}{\text{x+1}}$

That is,

$\frac{\text{dy}}{\text{dx}}=\frac{\text{-1}}{\text{x+1}}$                                                     …… (1)

Again, differentiating both sides with respect to $\text{x}$ gives

$\begin{array}{rl}& \frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{=}\frac{\text{d}}{\text{dx}}\text{=}\left(\frac{\text{1}}{\text{x+1}}\right)\text{=-}\left(\frac{\text{-1}}{{\text{(x+1)}}^{\text{2}}}\right)\text{=}\frac{\text{1}}{{\text{(x+1)}}^{\text{2}}}\\ & \Rightarrow \frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{=}{\left(\frac{\text{-1}}{\text{x+1}}\right)}^{\text{2}}\end{array}$

$\Rightarrow \frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{=}{\left(\frac{\text{dy}}{\text{dx}}\right)}^{\text{2}}$, using the equation (1).

Thus, it is proved that $\frac{{\text{d}}^{\text{2}}\text{y}}{\text{d}{\text{x}}^{\text{2}}}\text{=}{\left(\frac{\text{dy}}{\text{dx}}\right)}^{\text{2}}$.

17. If $\mathbf{y}\mathbf{=}\mathbf{(}\mathbf{ta}{\mathbf{n}}^{\mathbf{-}\mathbf{1}}\mathbf{x}{\mathbf{)}}^{\mathbf{2}}$, show that ${\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}\mathbf{)}}^{\mathbf{2}}{\mathbf{y}}_{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{x}\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{1}\mathbf{)}{\mathbf{y}}_{\mathbf{1}}\mathbf{=}\mathbf{2}$.

Ans:  

The given equations are $\text{y=(ta}{\text{n}}^{\text{-1}}\text{x}{\text{)}}^{\text{2}}$.

Then, differentiating both sides with respect to $\text{x}$ gives

$$\begin{array}{rl}& {\text{y}}_{\text{1}}\text{=2ta}{\text{n}}^{\text{-1}}\text{x}\frac{\text{d}}{\text{dx}}\text{(ta}{\text{n}}^{\text{-1}}\text{x)}\\ & \Rightarrow {\text{y}}_{\text{1}}\text{=2ta}{\text{n}}^{\text{-1}}\text{x }\times \frac{\text{1}}{\text{1+}{\text{x}}^{\text{2}}}\\ & \Rightarrow \text{(1+}{\text{x}}^{\text{2}}\text{)}{\text{y}}_{\text{1}}\text{=2ta}{\text{n}}^{\text{-1}}\text{x}\end{array}$$

Again, differentiating both sides with respect to $\text{x}$ gives

$\begin{array}{rl}& \text{(1+}{\text{x}}^{\text{2}}\text{)}{\text{y}}_{\text{2}}\text{+2x}{\text{y}}_{\text{1}}\text{=2}\left(\frac{\text{1}}{\text{1+}{\text{x}}^{\text{2}}}\right)\\ & \Rightarrow \text{(1+}{\text{x}}^{\text{2}}\text{)}{\text{y}}_{\text{2}}\text{+2x(1+}{\text{x}}^{\text{2}}\text{)}{\text{y}}_{\text{1}}\text{=2}\end{array}$

Thus, it has been proved that $\text{(1+}{\text{x}}^{\text{2}}\text{)}{\text{y}}_{\text{2}}\text{+2x(1+}{\text{x}}^{\text{2}}\text{)}{\text{y}}_{\text{1}}\text{=2}$.

Conclusion

Exercise 5.7 of Chapter 5 in Class 12 Maths focuses on second order derivatives, a crucial concept for understanding the behavior of functions. It is important to focus on accurately calculating second order derivatives. Regular practice of these problems will enhance your ability to analyze functions effectively. Vedantu's solutions provide detailed, step-by-step explanations to help you master these concepts, ensuring you are well-prepared for your exams. By understanding and practicing these key ideas, you'll build a solid foundation in calculus, which is essential for success in higher mathematics and various applications.

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