Question
Find $\frac{\text{dy}}{\text{ dx}} $in the following:
$\text{y}=\sin^{-1}\Bigg(\frac{2\text{x}}{1+\text{x}^{2}}\Bigg)$

Answer

The given relationship is $\text{y}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^{2}}\Big)$
$\text{y}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^{2}}\Big)$
$\Rightarrow \sin\text{y}=\frac{2\text{x}}{1+\text{x}^{2}}$
Differenting this relationship with respect to x, we obtain
$\frac{\text{d}}{\text{dx}}(\sin\text{y})=\frac{\text{d}}{\text{dx}}​​​​\Big(\frac{2\text{x}}{1+\text{x}^{2}}\Big)$
$\Rightarrow\cos\text{y}\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(\frac{2\text{x}}{1+\text{x}^{2}}\Big) ...(\text{i})$
The function$\frac{2\text{x}}{1+\text{x}^{2}}$, is of the from of $\frac{\text{u}}{\text{v}}.$
Therefore, by quotient rule, we obtain
$\frac{\text{d}}{\text{dx}}\Big(\frac{2\text{x}}{1+\text{x}^{2}}\Big)=\frac{(1+\text{x}^{2}).\frac{\text{d}}{\text{dx}}(2\text{x})-2\text{x}.\frac{\text{d}}{\text{dx}}(1 +\text{x}^{2})}{(1 +\text{x}^{2})^2}$
$=\frac{(1+\text{x}^{2}).2-2\text{x}.[0+2\text{x}]}{(1 +\text{x}^{2})^2}$$=\frac{2 + 2\text{x}^{2}-4\text{x}^{2}}{(1 +\text{x}^{2})^2}=\frac{2(1-\text{x}^{2})}{(1+\text{x}^{2})^2} ...(\text{ii})$
Also, $ \sin\text{y}=\frac{2\text{x}}{1+\text{x}^{2}}$
$\Rightarrow\cos\text{y}=\sqrt{1-\sin^{2}\text{y}}=\sqrt{1-\Big(\frac{2\text{x}}{1+\text{x}^{2}}\Big)^2}=\sqrt{\frac{(1+ \text{x}^{2})^2-4\text{x}^{2}}{(1 +\text{x}^{2})^{2}}}$
$=\sqrt{\frac{1+\text{x}^4+2\text{x}^2-4\text{x}^2}{(1+\text{x}^2)^2}}=\sqrt{\frac{(1-\text{x}^2)^2}{(1+\text{x}^2)^2}}$
$=\sqrt{\frac{(1- \text{x}^{2})^{2}}{(1 +\text{x}^{2})^{2}}}=\frac{1-\text{x}^{2}}{1+\text{x}^{2}} ...\text{(iii)}$
Form(1), (2), and (3), we obtain
$\frac{1-\text{x}^{2}}{1+\text{x}^{2}}\times\frac{\text{dy}}{\text{dx}}=\frac{2(1-\text{x}^{2})}{(1 +\text{x}^{2})^{2}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=\frac{2}{1+\text{x}^{2}}$

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