# Find the value

Question:

Find $\frac{d y}{d x}$, When $y=\frac{\sin x+x^{2}}{\cot 2 x}$

Solution:

Let $y=\frac{\sin x+x^{2}}{\cot 2 x}, u=\sin x+x^{2}, v=\cot 2 x$

Formula:

$\frac{\mathrm{d}(\sin \mathrm{x})}{\mathrm{dx}}=\cos \mathrm{x}, \frac{\mathrm{d}\left(\mathrm{x}^{\mathrm{n}}\right)}{\mathrm{dx}}=\mathrm{n} \times \mathrm{x}^{\mathrm{n}-1}$ and $\frac{\mathrm{d}(\cot \mathrm{x})}{\mathrm{dx}}=-\operatorname{cosec}^{2} \mathrm{x}$

According to the quotient rule of differentiation

If $y=\frac{u}{v}$

$\mathrm{dy} / \mathrm{dx}=\frac{\mathrm{v} \times \frac{\mathrm{du}}{\mathrm{dx}}-\mathrm{u} \times \frac{\mathrm{dv}}{\mathrm{dx}}}{\mathrm{v}^{2}}$

$=\frac{(\cot 2 x) \times(\cos x+2 x)-\left(\sin x+x^{2}\right) \times\left(-2 \operatorname{cosec}^{2} 2 x\right)}{(\cot 2 x)^{2}}$

$=\frac{\cot 2 x \cos x+2 x \cot 2 x+2 \operatorname{cosec}^{2} 2 x \sin x+2 x^{2} \operatorname{cosec}^{2} 2 x}{(\operatorname{cosec} 2 x)^{2}}$

$=\frac{\cot 2 x(\cos x+2 x)+2 \operatorname{cosec}^{2} 2 x\left(\sin x+x^{2}\right)}{(\operatorname{cosec} 2 x)^{2}}$

$=\frac{2 \operatorname{cosec}^{2} 2 x\left(\sin x+x^{2}\right)}{(\operatorname{cosec} 2 x)^{2}}+\frac{\cot 2 x(\cos x+2 x)}{(\operatorname{cosec} 2 x)^{2}}$

$=\frac{2\left(\sin x+x^{2}\right)}{1}+\frac{\cos 2 x(\cos x+2 x)}{\sin 2 x \frac{1}{\sin ^{2} 2 x}}$

$=2\left(\sin x+x^{2}\right)+\cos 2 x \sin 2 x(\cos x+2 x)$