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Question
Differentiate \[\frac{x^2 \left( 1 - x^2 \right)}{\cos 2x}\] ?
Solution
\[\text{Let } y = \frac{x^2 \left( 1 - x^2 \right)^3}{\cos 2x}\]
\[\Rightarrow \frac{d y}{d x} = \frac{\cos2x\frac{d}{dx}\left\{ x^2 \left( 1 - x^2 \right)^3 \right\} - x^2 \left( 1 - x^2 \right)^3 \frac{d}{dx}\cos2x}{\cos^2 2x} \]
\[ = \frac{\cos2x\left\{ x^2 \frac{d}{dx} \left( 1 - x^2 \right)^3 + \left( 1 - x^2 \right)^3 \frac{d}{dx}\left( x^2 \right) \right\} - x^2 \left( 1 - x^2 \right)^3 \left( - 2\sin2x \right)}{\cos^2 2x}\]
\[ = \frac{\cos2x\left\{ - 6 x^3 \left( 1 - x^2 \right)^2 + \left( 1 - x^2 \right)^3 2x \right\} + 2 x^2 \left( 1 - x^2 \right)^3 \sin2x}{\cos^2 2x}\]
\[ = \frac{2x \left( 1 - x^2 \right)^2}{\cos2x} - \frac{6 x^3 \left( 1 - x^2 \right)^2}{\cos2x} + \frac{2 x^2 \left( 1 - x^2 \right)^3 \sin2x}{\cos^2 2x}\]
\[ = 2x\left( 1 - x^2 \right)\sec2x\left\{ 1 - 4 x^2 + x\left( 1 - x^2 \right)\tan2x \right\}\]
\[So, \frac{d}{dx}\left\{ \frac{x^2 \left( 1 - x^2 \right)^3}{\cos2x} \right\} = 2x\left( 1 - x^2 \right)\sec2x\left\{ 1 - 4 x^2 + x\left( 1 - x^2 \right)\tan2x \right\}\]
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