Differentiate the following functions with respect to x :

Solution:

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

On differentiating $y$ with respect to $x$, we get

$\frac{d y}{d x}=\frac{d}{d x}\left[\frac{x^{2}\left(1-x^{2}\right)^{3}}{\cos 2 x}\right]$

Recall that $\left(\frac{\mathrm{u}}{\mathrm{v}}\right)^{\prime}=\frac{\mathrm{vu}^{\prime}-\mathrm{uv}^{\prime}}{\mathrm{v}^{2}}$ (quotient rule)

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

We have (uv)' $=$ vu' $+$ uv' (product rule)

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}$

$=\frac{\cos 2 x\left[\left(1-x^{2}\right)^{3} \frac{d}{d x}\left(x^{2}\right)+x^{2} \frac{d}{d x}\left\{\left(1-x^{2}\right)^{3}\right\}\right]-x^{2}\left(1-x^{2}\right)^{3} \frac{d}{d x}(\cos 2 x)}{\cos ^{2} 2 x}$

We know $\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}$ and $\frac{d}{d x}(\cos x)=-\sin x$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}$

$=\frac{\cos 2 x\left[\left(1-x^{2}\right)^{3}(2 x)+x^{2}\left\{3\left(1-x^{2}\right)^{2} \frac{d}{d x}\left(1-x^{2}\right)\right\}\right]-x^{2}\left(1-x^{2}\right)^{3}\left(-\sin 2 x \frac{d}{d x}(2 x)\right)}{\cos ^{2} 2 x}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}$

$=\frac{\cos 2 x\left[2 x\left(1-x^{2}\right)^{3}+3 x^{2}\left(1-x^{2}\right)^{2}\left\{\frac{d}{d x}(1)-\frac{d}{d x}\left(x^{2}\right)\right\}\right]+2 x^{2}\left(1-x^{2}\right)^{3} \sin 2 x \frac{d}{d x}(x)}{\cos ^{2} 2 x}$

However, $\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}$ and derivative of a constant is 0 .

$\Rightarrow \frac{d y}{d x}=\frac{\cos 2 x\left[2 x\left(1-x^{2}\right)^{3}+3 x^{2}\left(1-x^{2}\right)^{2}\{0-2 x\}\right]+2 x^{2}\left(1-x^{2}\right)^{3} \sin 2 x \times 1}{\cos ^{2} 2 x}$

$\Rightarrow \frac{d y}{d x}=\frac{\cos 2 x\left[2 x\left(1-x^{2}\right)^{3}+3 x^{2}\left(1-x^{2}\right)^{2}(-2 x)\right]+2 x^{2}\left(1-x^{2}\right)^{3} \sin 2 x}{\cos ^{2} 2 x}$

$\Rightarrow \frac{d y}{d x}=\frac{\cos 2 x\left[2 x\left(1-x^{2}\right)^{3}-6 x^{3}\left(1-x^{2}\right)^{2}\right]+2 x^{2}\left(1-x^{2}\right)^{3} \sin 2 x}{\cos ^{2} 2 x}$

$\Rightarrow \frac{d y}{d x}=\frac{2 x\left(1-x^{2}\right)^{3} \cos 2 x-6 x^{3}\left(1-x^{2}\right)^{2} \cos 2 x+2 x^{2}\left(1-x^{2}\right)^{3} \sin 2 x}{\cos ^{2} 2 x}$

$\Rightarrow \frac{d y}{d x}=\frac{2 x\left(1-x^{2}\right)^{3} \cos 2 x}{\cos ^{2} 2 x}-\frac{6 x^{3}\left(1-x^{2}\right)^{2} \cos 2 x+}{\cos ^{2} 2 x}+\frac{2 x^{2}\left(1-x^{2}\right)^{3} \sin 2 x}{\cos ^{2} 2 x}$

$\Rightarrow \frac{d y}{d x}=\frac{2 x\left(1-x^{2}\right)^{3}}{\cos 2 x}-\frac{6 x^{3}\left(1-x^{2}\right)^{2}}{\cos 2 x}+\frac{2 x^{2}\left(1-x^{2}\right)^{3} \sin 2 x}{\cos 2 x \times \cos 2 x}$

$\Rightarrow \frac{d y}{d x}=\frac{2 x\left(1-x^{2}\right)^{3}}{\cos 2 x}-\frac{6 x^{3}\left(1-x^{2}\right)^{2}}{\cos 2 x}+\frac{2 x^{2}\left(1-x^{2}\right)^{3} \tan 2 x}{\cos 2 x}$

$\Rightarrow \frac{d y}{d x}=\frac{2 x\left(1-x^{2}\right)^{2}}{\cos 2 x}\left[\left(1-x^{2}\right)-3 x^{2}+x\left(1-x^{2}\right) \tan 2 x\right]$

$\Rightarrow \frac{d y}{d x}=\frac{2 x\left(1-x^{2}\right)^{2}}{\cos 2 x}\left[1-4 x^{2}+x\left(1-x^{2}\right) \tan 2 x\right]$

$\therefore \frac{d y}{d x}=2 x\left(1-x^{2}\right)^{2} \sec 2 x\left[1-4 x^{2}+x\left(1-x^{2}\right) \tan 2 x\right]$

Thus, $\frac{d}{d x}\left[\frac{x^{2}\left(1-x^{2}\right)^{3}}{\cos 2 x}\right]=2 x\left(1-x^{2}\right)^{2} \sec 2 x\left[1-4 x^{2}+x\left(1-x^{2}\right) \tan 2 x\right]$