x = e ( + ),y = e - ( - )

and y = e - ( - ) therefore, d = d = e d ( + ) + ( + ) d e = e ( 1 - 2 ) + ( + ) e = e ( 1 - 2 + + ) = e ( - 1 + 3 + 2 ) ……(i) and d = d = e - d ( - ) + d e - ( - ) = e - ( 1 + 2 ) + ( - ) e - d ( - ) = e - = e - ……(ii) therefore, dx = dx/d = ( + 1 - 3 + 2 ) e ( - 1 + 3 + 2 ) = e - 2 ( + 2 + + 1 3 + 2 + - 1 ) dx = e - 2 ( + 2 + + 1 3 + 2 + - 1 )

class 12 maths continuity and differentiability

$x = {e^\theta }\left( {\theta + \frac{1}{\theta }} \right),y = {e^{ - \theta }}\left( {\theta - \frac{1}{\theta }} \right)$

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#Continuity and Differentiability #Exemplar #SA #Class 12 Maths

📘 Continuity and Differentiability NCERT Exemp. Ex.5.3 ,Q.45,Page 110 SA

$x = {e^\theta }\left( {\theta + \frac{1}{\theta }} \right),y = {e^{ - \theta }}\left( {\theta - \frac{1}{\theta }} \right)$

Official Solution

VVidaara Team ✓ Verified solution NCERT & Exemplar

and $y = {e^{ - \theta }}\left( {\theta - \frac{1}{\theta }} \right)$

therefore,$\frac{{dx}}{{d\theta }} = \frac{d}{{d\theta }}\left[ {{e^\theta } \cdot \left( {\theta + \frac{1}{\theta }} \right)} \right]$

$= {e^\theta } \cdot \frac{d}{{d\theta }}\left( {\theta + \frac{1}{\theta }} \right) + \left( {\theta + \frac{1}{\theta }} \right) \cdot \frac{d}{{d\theta }}{e^\theta }$

$= {e^\theta }\left( {1 - \frac{1}{{{\theta ^2}}}} \right) + \left( {\theta + \frac{1}{\theta }} \right){e^\theta }$

$= {e^\theta }\left( {1 - \frac{1}{{{\theta ^2}}} + \theta + \frac{1}{\theta }} \right)$

$= {e^\theta }\left( {\frac{{{\theta ^2} - 1 + {\theta ^3} + \theta }}{{{\theta ^2}}}} \right)$ ……(i)

and $\frac{{dy}}{{d\theta }} = \frac{d}{{d\theta }}\left[ {{e^{ - \theta }} \cdot \left( {\theta - \frac{1}{\theta }} \right)} \right]$

$= {e^{ - \theta }} \cdot \frac{d}{{d\theta }}\left( {\theta - \frac{1}{\theta }} \right) + \frac{d}{{d\theta }}{e^{ - \theta }}\left( {\theta - \frac{1}{\theta }} \right)$

$= {e^{ - \theta }}\left( {1 + \frac{1}{{{\theta ^2}}}} \right) + \left( {\theta - \frac{1}{\theta }} \right){e^{ - \theta }} \cdot \frac{d}{{d\theta }}( - \theta )$

$= {e^{ - \theta }}\left[ {\frac{{{\theta ^2} + 1}}{{{\theta ^2}}} - \frac{{{\theta ^2} - 1}}{\theta }} \right] = {e^{ - \theta }}\left[ {\frac{{{\theta ^2} + 1 - {\theta ^3} + \theta }}{{{\theta ^2}}}} \right]$ ……(ii)

therefore,$\frac{{dy}}{{dx}} = \frac{{dy/d\theta }}{{dx/d\theta }} = \frac{{{e^{ - \theta }}\left( {\frac{{{\theta ^2} + 1 - {\theta ^3} + \theta }}{{{\theta ^2}}}} \right)}}{{{e^\theta }\left( {\frac{{{\theta ^2} - 1 + {\theta ^3} + \theta }}{{{\theta ^2}}}} \right)}}$

$= {e^{ - 2\theta }}\left( {\frac{{ - {\theta ^3} + {\theta ^2} + \theta + 1}}{{{\theta ^3} + {\theta ^2} + \theta - 1}}} \right)$

$\Rightarrow \frac{dy}{dx} = {e^{ - 2\theta }}\left( {\frac{{ - {\theta ^3} + {\theta ^2} + \theta + 1}}{{{\theta ^3} + {\theta ^2} + \theta - 1}}} \right)$

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