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Find D Y D X Y = ( X 2 − 1 ) 3 ( 2 X − 1 ) √ ( X − 3 ) ( 4 X − 1 ) ? - CBSE (Arts) Class 12 - Mathematics

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Question

find  \[\frac{dy}{dx}\]  \[y = \frac{\left( x^2 - 1 \right)^3 \left( 2x - 1 \right)}{\sqrt{\left( x - 3 \right) \left( 4x - 1 \right)}}\] ?

 

Solution

\[\text{ We have, y } = \frac{\left( x^2 - 1 \right)^3 \left( 2x - 1 \right)}{\sqrt{\left( x - 3 \right)\left( 4x - 1 \right)}} . . . \left( i \right)\]

\[ \Rightarrow y = \frac{\left( x^2 - 1 \right)^3 \left( 2x - 1 \right)}{\left( x - 3 \right)^\frac{1}{2} \left( 4x - 1 \right)^\frac{1}{2}}\]

Taking log on both sides,

\[\log y = \log\left[ \frac{\left( x^2 - 1 \right)^3 \left( 2x - 1 \right)}{\left( x - 3 \right)^\frac{1}{2} \left( 4x - 1 \right)^\frac{1}{2}} \right]\]
\[ \Rightarrow \log y = \log \left( x^2 - 1 \right)^3 + \log\left( 2x - 1 \right) - \log \left( x - 3 \right)^\frac{1}{2} - \log \left( 4x - 1 \right)^\frac{1}{2} \]
\[ \Rightarrow \log y = 3 \log\left( x^2 - 1 \right) + \log\left( 2x - 1 \right) - \frac{1}{2}\log\left( x - 3 \right) - \frac{1}{2}\log\left( 4x - 1 \right)\]

Differentiating with respect to x using chain rule,

\[\frac{1}{y}\frac{dy}{dx} = 3\frac{d}{dx}\left\{ \log\left( x^2 - 1 \right) \right\} + \frac{d}{dx}\left\{ \log\left( 2x - 1 \right) \right\} - \frac{1}{2}\frac{d}{dx}\left\{ \log\left( x - 3 \right) \right\} - \frac{1}{2}\left\{ \log\left( 4x - 1 \right) \right\}\]
\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = 3\left( \frac{1}{x^2 - 1} \right)\frac{d}{dx}\left( x^2 - 1 \right) + \frac{1}{\left( 2x - 1 \right)}\frac{d}{dx}\left( 2x - 1 \right) - \frac{1}{2}\left( \frac{1}{x - 3} \right)\frac{d}{dx}\left( x - 3 \right) - \frac{1}{2}\frac{1}{\left( 4x - 1 \right)}\frac{d}{dx}\left( 4x - 1 \right)\]
\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = 3\left( \frac{1}{x^2 - 1} \right)\left( 2x \right) + \frac{1}{2x - 1}\left( 2 \right) - \frac{1}{2}\left( \frac{1}{x - 3} \right)\left( 1 \right) - \frac{1}{2}\left( \frac{1}{4x - 1} \right)\left( 4 \right)\]
\[ \Rightarrow \frac{1}{y}\frac{dy}{dx} = \left[ \frac{6x}{x^2 - 1} + \frac{2}{2x - 1} - \frac{1}{2\left( x - 3 \right)} - \frac{2}{4x - 1} \right]\]
\[ \Rightarrow \frac{dy}{dx} = y\left[ \frac{6x}{x^2 - 1} + \frac{2}{2x - 1} - \frac{1}{2\left( x - 3 \right)} - \frac{2}{4x - 1} \right]\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\left( x^2 - 1 \right)^3 \left( 2x - 1 \right)}{\sqrt{\left( x - 3 \right)\left( 4x - 1 \right)}}\left[ \frac{6x}{x^2 - 1} + \frac{2}{2x - 1} - \frac{1}{2\left( x - 3 \right)} - \frac{2}{4x - 1} \right] \left[ \text{ using equation} \left( i \right) \right]\]

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Solution Find D Y D X Y = ( X 2 − 1 ) 3 ( 2 X − 1 ) √ ( X − 3 ) ( 4 X − 1 ) ? Concept: Simple Problems on Applications of Derivatives.
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