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Question
If \[f'\left( x \right) = \sqrt{2 x^2 - 1} \text { and y } = f \left( x^2 \right)\] then find \[\frac{dy}{dx} \text { at } x = 1\] ?
Solution
\[\text { We have,} f'\left( x \right) = \sqrt{2 x^2 - 1}\]
\[\text {and y } = f\left( x^2 \right)\]
\[\Rightarrow \frac{dy}{dx} = \frac{d}{dx}f\left( x^2 \right)\]
\[ \Rightarrow \frac{dy}{dx} = f'\left( x^2 \right)\frac{d}{dx}\left( x^2 \right)\]
\[ \Rightarrow \frac{dy}{dx} = f'\left( x^2 \right) \times 2x\]
\[ \Rightarrow \frac{dy}{dx} = 2xf'\left( x^2 \right)\]
\[\text { Putting x } = 1, \text { we get }, \]
\[\frac{dy}{dx} = 2\left( 1 \right)f'\left( 1^2 \right)\]
\[ \Rightarrow \frac{dy}{dx} = 2 \times f'\left( 1 \right)\]
\[ \Rightarrow \frac{dy}{dx} = 2 \times 1 \left[ \because f'\left( 1 \right) = \sqrt{2 \left( 1 \right)^2 - 1} = \sqrt{2 - 1} = 1 \right]\]
\[ \Rightarrow \frac{dy}{dx} = 2\]
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