Question

Differentiate

$\frac{x^2{\left(1-x^2\right)}^3}{\cos 2x}$

Answer 1

$\mathrm{Let}y=\frac{x^2{\left(1-x^2\right)}^3}{\cos 2x}$
$$\begin{gathered} \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=\frac{\cos 2x{\displaystyle \frac{d}{dx}}\left\{x^2{\left(1-x^2\right)}^3\right\}-x^2{\left(1-x^2\right)}^3{\displaystyle \frac{d}{dx}}\cos 2x}{{\cos }^{2}2x} \\ =\frac{\cos 2x\left\{x^2{\displaystyle \frac{d}{dx}}{\left(1-x^2\right)}^3+{\left(1-x^2\right)}^3{\displaystyle \frac{d}{dx}}\left(x^2\right)\right\}-x^2{\left(1-x^2\right)}^3\left(-2\sin 2x\right)}{{\cos }^{2}2x} \\ =\frac{\cos 2x\left\{-6x^3{\left(1-x^2\right)}^2+{\left(1-x^2\right)}^{3}2x\right\}+2x^2{\left(1-x^2\right)}^3\sin 2x}{{\cos }^{2}2x} \\ =\frac{2x{\left(1-x^2\right)}^2}{\cos 2x}-\frac{6x^3{\left(1-x^2\right)}^2}{\cos 2x}+\frac{2x^2{\left(1-x^2\right)}^3\sin 2x}{{\cos }^{2}2x} \\ =2x\left(1-x^2\right)\sec 2x\left\{1-4x^2+x\left(1-x^2\right)\tan 2x\right\} \\ \mathrm{So},\frac{d}{dx}\left\{\frac{x^2{\left(1-x^2\right)}^3}{\cos 2x}\right\}=2x\left(1-x^2\right)\sec 2x\left\{1-4x^2+x\left(1-x^2\right)\tan 2x\right\} \end{gathered}$$