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CBSE Class 12 Mathematics Chapter 5 Continuity and Differentiability – NCERT Solutions 2025-26

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Download Free PDF of Continuity and Differentiability Exercise 5.7 for Class 12 Maths

If you’re preparing for Class 12 CBSE board exams, mastering NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.7 can make a noticeable difference to your result. This crucial exercise in Continuity and Differentiability helps you strengthen concepts like the chain rule, implicit differentiation, and derivatives of composite and trigonometric functions—topics consistently valued by the board and worth up to 9 marks in calculus.

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With reliable answers developed by experienced CBSE educators, you get the clarity and accuracy needed for high-stakes exams. For structured syllabus reference, use the Class 12 Maths syllabus as your official guide.

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Access Class 12 Maths NCERT Chapter 5 Continuity and Differentiability Exercise 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. $\mathbf{y=}{{\mathbf{x}}^{\mathbf{20}}}$.

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

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

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

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


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

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

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

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


6. $\mathbf{y=}{{\mathbf{e}}^{\mathbf{x}}}\mathbf{sin5x}$

Ans: 

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

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

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

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

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

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

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

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

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

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

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

\[\begin{align} & \text{=36}{{\text{e}}^{\text{6x}}}\text{cos3x-18}{{\text{e}}^{\text{6x}}}\text{sin3x-3}\left[ \text{sin3x }\!\!\times\!\!\text{ }{{\text{e}}^{\text{6x}}}\text{ }\!\!\times\!\!\text{ 6+}{{\text{e}}^{\text{6x}}}\text{ }\!\!\times\!\!\text{ cos3x-3} \right] \\ & \text{=36}{{\text{e}}^{\text{6x}}}\text{cos3x-18}{{\text{e}}^{\text{6x}}}\text{sin3x-18}{{\text{e}}^{\text{6x}}}\text{sin3x-9}{{\text{e}}^{\text{6x}}}\text{cos3x} \\ \end{align}\]

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


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

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

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

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

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

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

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

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

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

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

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

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

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

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

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 $\mathbf{y=co}{{\mathbf{s}}^{\mathbf{-1}}}\mathbf{x}$, 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

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

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

                                                                                                                       

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

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

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

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

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

\[\begin{align} & {{\text{y}}_{\text{1}}}\text{=3 }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ cos(logx) }\!\!]\!\!\text{ +4 }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{ }\!\![\!\!\text{ sin(logx) }\!\!]\!\!\text{ } \\ & \text{=3 }\!\!\times\!\!\text{ }\left[ \text{-sin(logx) }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(logx)} \right]\text{+4 }\!\!\times\!\!\text{ }\left[ \text{cos(logx) }\!\!\times\!\!\text{ }\frac{\text{d}}{\text{dx}}\text{(logx)} \right] \\ \end{align}\]

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

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

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

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

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

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

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

Now, substituting the derivatives ${{\text{y}}_{1}}$ ,${{\text{y}}_{2}}$ and $\text{y}$ into the LHS of the equation (2) gives

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

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

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

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

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

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

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

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

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

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

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

$\text{=49y}$, 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 ${{\mathbf{e}}^{\mathbf{y}}}\mathbf{(x+1)=1}$, 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

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

That is,

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

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

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

$\Rightarrow \frac{{{\text{d}}^{\text{2}}}\text{y}}{\text{d}{{\text{x}}^{\text{2}}}}\text{=}{{\left( \frac{\text{dy}}{\text{dx}} \right)}^{\text{2}}}$, 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 $\mathbf{y=(ta}{{\mathbf{n}}^{\mathbf{-1}}}\mathbf{x}{{\mathbf{)}}^{\mathbf{2}}}$, show that ${{\mathbf{(}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{+1)}}^{\mathbf{2}}}{{\mathbf{y}}_{\mathbf{2}}}\mathbf{+2x(}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{+1)}{{\mathbf{y}}_{\mathbf{1}}}\mathbf{=2}$.

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

\[\begin{align} & {{\text{y}}_{\text{1}}}\text{=2ta}{{\text{n}}^{\text{-1}}}\text{x}\frac{\text{d}}{\text{dx}}\text{(ta}{{\text{n}}^{\text{-1}}}\text{x)} \\ & \Rightarrow {{\text{y}}_{\text{1}}}\text{=2ta}{{\text{n}}^{\text{-1}}}\text{x }\!\!\times\!\!\text{ }\frac{\text{1}}{\text{1+}{{\text{x}}^{\text{2}}}} \\ & \Rightarrow \text{(1+}{{\text{x}}^{\text{2}}}\text{)}{{\text{y}}_{\text{1}}}\text{=2ta}{{\text{n}}^{\text{-1}}}\text{x} \\ \end{align}\]

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

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

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.


Class 12 Maths Chapter 5: Exercises Breakdown

S.No.

Chapter 5 - Continuity and Differentiability Exercises in PDF Format

1

Class 12 Maths Chapter 5 Exercise 5.1 - 34 Questions & Solutions (10 Short Answers, 24 Long Answers)

2

Class 12 Maths Chapter 5 Exercise 5.2 - 10 Questions & Solutions (2 Short Answers, 8 Long Answers)

3

Class 12 Maths Chapter 5 Exercise 5.3 - 15 Questions & Solutions (9 Short Answers, 6 Long Answers)

4

Class 12 Maths Chapter 5 Exercise 5.4 - 10 Questions & Solutions (5 Short Answers, 5 Long Answers)

5

Class 12 Maths Chapter 5 Exercise 5.5 - 18 Questions & Solutions (4 Short Answers, 14 Long Answers)

6

Class 12 Maths Chapter 5 Exercise 5.6 - 11 Questions & Solutions (7 Short Answers, 4 Long Answers)

7

Class 12 Maths Chapter 5 Miscellaneous Exercise - 22 Questions & Solutions



CBSE Class 12 Maths Chapter 5 Other Study Materials



NCERT Solutions for Class 12 Maths | Chapter-wise List

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FAQs on CBSE Class 12 Mathematics Chapter 5 Continuity and Differentiability – NCERT Solutions 2025-26

1. How do I apply the chain rule in question 3 of Exercise 5.7?

The chain rule in Class 12 Maths Chapter 5 Exercise 5.7 lets you find derivatives of composite functions stepwise.
To apply the chain rule:

  • Identify the outer function and inner function in the given expression.
  • Differenti­ate the outer function, keeping the inner function as it is.
  • Multiply by the derivative of the inner function.
  • Simplify thoroughly and match each step to the NCERT solution pattern.
Practice this approach for all composite expressions found in the exercise, using key terms like 'chain rule', 'derivative of composite function', and 'CBSE stepwise solution'.

2. Which formulas are most important for Continuity and Differentiability in Class 12 CBSE?

The most important formulas for Chapter 5 are those frequently used in board and NCERT exercises.
Key formulas include:

  • Chain Rule: d/dx [f(g(x))] = f'(g(x)) · g'(x)
  • Derivative of inverse trigonometric functions: e.g., d/dx [sin-1x] = 1/√(1-x2)
  • Implicit Differentiation technique
  • Derivatives of exponential and logarithmic functions
  • Standard derivatives for all trigonometric functions
Memorise and revise these using a formula sheet and by practicing solved examples in Exercise 5.7 for high marks on CBSE exams.

3. What is the stepwise process for solving implicit differentiation questions in Ex 5.7?

To solve implicit differentiation questions in NCERT Exercise 5.7:
Follow these steps:

  1. Differenti­ate both sides of the equation with respect to x.
  2. Treat y as an implicit function: apply chain rule when differentiating terms with y.
  3. Whenever you differentiate y, multiply the result by dy/dx.
  4. Collect all terms containing dy/dx on one side of the equation and move other terms to the opposite side.
  5. Solve for dy/dx to get the required derivative.
This approach is essential for CBSE 'continuity and differentiability' problems, and ensures stepwise scoring for the board exam.

4. Can I download the solutions for Exercise 5.7 as a PDF for offline study?

Yes, you can download the complete NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.7 as a PDF for offline revision.
Benefits of downloading the PDF:

  • Access stepwise, CBSE-aligned answers anytime
  • Easy last-minute revision for board exams
  • Better clarity with highlighted formulas and detailed explanations
This ensures you can prepare for continuity and differentiability from trusted sources, without internet access.

5. How does Vedantu ensure all solutions are CBSE board exam-compliant?

Vedantu guarantees board exam compliance in Class 12 Maths Ex 5.7 solutions through several measures:

  • Mapping all solutions to the latest CBSE/NCERT syllabus (2025)
  • Stepwise answers as per CBSE marking scheme
  • Peer review by expert Mathematics faculty and recent toppers
  • Regular updates for accuracy and syllabus changes
All content is trusted for conceptual clarity, error-free learning, and board exam preparation per the latest curriculum.

6. How to score maximum marks in derivatives chapter for the board exam?

To score full marks in the derivatives chapter (Continuity and Differentiability):
Follow these steps:

  • Practice all NCERT solved examples and Exercise 5.7 solutions step-by-step
  • Memorise critical formulas and understand their application
  • Write each derivative and algebraic step clearly—no skips
  • Use proper mathematical reasoning for chain rule and implicit differentiation
  • Review CBSE previous years' questions for exam-trending patterns
Consistency with revision notes and a focus on error-free working boost marks in this high-weightage unit.

7. What topics are covered in NCERT Class 12 Maths Chapter 5 Exercise 5.7?

NCERT Class 12 Maths Chapter 5 Exercise 5.7 covers key differentiability topics required for board exams:

  • Chain rule application for composite functions
  • Implicit differentiation
  • Derivatives of inverse trigonometric and exponential functions
  • Error-checking and stepwise algebraic manipulation
These concepts are essential for mastering continuity and differentiability, as outlined by the latest 2025 CBSE syllabus.

8. How many exercises are in continuity and differentiability class 12?

There are typically 7 main exercises plus a miscellaneous exercise in Chapter 5 (Continuity and Differentiability) of NCERT Class 12 Maths.

  • Each exercise focuses on a specific concept: basic derivatives, chain rule, implicit differentiation, and more.
  • Exercise 5.7 practises advanced differentiation methods crucial for the board exam.
This structure ensures comprehensive coverage of all calculus fundamentals prescribed by CBSE.

9. Why use stepwise NCERT Solutions for Exercise 5.7 instead of shortcuts?

Using stepwise NCERT Solutions for Exercise 5.7 ensures you:

  • Gain clear understanding of core concepts like chain rule and implicit differentiation
  • Match every mark in the CBSE board exam's scheme
  • Reduce errors and improve answer writing skills
  • Score higher as steps are explicitly marked, not just final answers
This method is recommended by teachers for board success in continuity and differentiability.

10. Which CBSE board exams and entrance tests is Continuity and Differentiability relevant for?

Continuity and Differentiability (Class 12, Chapter 5) is important for:

  • CBSE Board exam (high calculus weightage, regular board questions)
  • NEET (medical entrance, calculus section)
  • JEE Main (engineering entrance, concept and application based)
Solutions from this chapter are also valuable for quick revision before all Class 12 maths assessments and subject Olympiads.

11. What is the LCD in math class 12?

In Class 12 Maths, LCD stands for Lowest Common Denominator. It is mainly used to add or subtract rational expressions by expressing them with the same denominator.

  • It is not the primary focus of Chapter 5 (Continuity and Differentiability), but knowing the LCD helps simplify expressions while differentiating fractions.
This topic may appear in earlier rational function problems or in algebra steps during differentiation.

12. How to check continuity class 12?

To check continuity in Class 12 Maths:

  • Find the left hand limit (LHL), right hand limit (RHL), and the function value at the point of interest.
  • If LHL = RHL = f(a), the function is continuous at x = a.
This method applies to all NCERT board exam questions on continuity, especially in the first part of Chapter 5.