Question

If a function is given by $y={{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)$, then find $\dfrac{dy}{dx}$.

Answer 1

At first suppose ${{\sin }^{-1}}x$ as f (x) and $\dfrac{2x}{1+{{x}^{2}}}$ as g (x). So we can write y as f (g (x)). Now for differentiation we will use the following rule, which is known as chain rule $\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)=f'\left( g\left( x \right) \right)\times g'\left( x \right)$, where $f'\left( g\left( x \right) \right)$ is differentiation of f (x) keeping g (x) constant and $g'\left( x \right)$ is differentiation of g (x) is irrespective of what f (x).

Complete step-by-step answer:
In the question we are given an expression of y which is ${{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)$ and we have to differentiate the function y with respect to x and find $\dfrac{dy}{dx}$.
Now we are asked to find $\left( \dfrac{dy}{dx} \right)$ which means we have to differentiate y with respect to x.
So let us consider two functions f (x) and g (x) where let f (x) be ${{\sin }^{-1}}x$ and g (x) be $\dfrac{2x}{1+{{x}^{2}}}$.
So, we can write,
${{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)$ or $f\left( g\left( x \right) \right)$.
Now we have to differentiate y with respect to x using the identity,
$\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)=f'\left( g\left( x \right)\times g'\left( x \right) \right)$
Here $f'\left( g\left( x \right) \right)$ means differentiating f (x) keeping g (x) constant here g’ (x) means differentiating g (x) independently irrespective of what f (x) is.

& \dfrac{d}{dx}\left( {{\sin }^{-1}}x \right)=\dfrac{1}{\sqrt{1-{{x}^{2}}}} \\\ & \dfrac{d}{dx}\left( \dfrac{2x}{1+{{x}^{2}}} \right)=2x\left( \dfrac{d}{dx}\left( \dfrac{1}{1+{{x}^{2}}} \right) \right)+\dfrac{1}{1+{{x}^{2}}}\left( \dfrac{d}{dx}\left( 2x \right) \right) \\\ \end{aligned}$$ Which we did by using formula, $$\dfrac{d}{dx}\left( h\left( x \right)k\left( x \right) \right)=\left\\{ \dfrac{d}{dx}\left( h\left( x \right) \right) \right\\}k\left( x \right)+\left\\{ \dfrac{d}{dx}\left( k\left( x \right) \right) \right\\}h\left( x \right)$$ $$\Rightarrow \dfrac{d}{dx}\left( \dfrac{2x}{1+{{x}^{2}}} \right)=\dfrac{-2x\times 2x}{{{\left( 1+{{x}^{2}} \right)}^{2}}}+\dfrac{1}{1+{{x}^{2}}}\times 2$$ $$\Rightarrow \dfrac{d}{dx}\left( \dfrac{2x}{1+{{x}^{2}}} \right)=\dfrac{-4{{x}^{2}}}{{{\left( 1+{{x}^{2}} \right)}^{2}}}+\dfrac{2}{\left( 1+{{x}^{2}} \right)}$$ $$\Rightarrow \dfrac{d}{dx}\left( \dfrac{2x}{1+{{x}^{2}}} \right)=\dfrac{2\left( 1+{{x}^{2}} \right)-4{{x}^{2}}}{{{\left( 1+{{x}^{2}} \right)}^{2}}}=\dfrac{2\left( 1-{{x}^{2}} \right)}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$$ So we get, $$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-{{\left( \dfrac{2x}{1+{{x}^{2}}} \right)}^{2}}}}\times \dfrac{2\left( 1-{{x}^{2}} \right)}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$$ Hence on simplification we get, $$\Rightarrow \dfrac{dy}{dx}=\dfrac{1+{{x}^{2}}}{\sqrt{{{\left( 1+{{x}^{2}} \right)}^{2}}-4{{x}^{2}}}}\times \dfrac{2\left( 1-{{x}^{2}} \right)}{{{\left( 1+{{x}^{2}} \right)}^{2}}}$$ We used the identity that, $${{\left( a-b \right)}^{2}}={{\left( a+b \right)}^{2}}-4ab$$, where a is 1 and b is $${{x}^{2}}$$. So, we get, $$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\sqrt{{{\left( 1-{{x}^{2}} \right)}^{2}}}}\times \dfrac{2\left( 1-{{x}^{2}} \right)}{\left( 1+{{x}^{2}} \right)}$$ $$\Rightarrow $$ $$\dfrac{dy}{dx}=\dfrac{1}{\left( 1-{{x}^{2}} \right)}\times \dfrac{2\left( 1-{{x}^{2}} \right)}{\left( 1+{{x}^{2}} \right)}$$ $$\Rightarrow $$ $$\dfrac{dy}{dx}=\dfrac{2}{\left( 1+{{x}^{2}} \right)}$$ Hence the value of $$\dfrac{dy}{dx}$$ is $$\dfrac{2}{\left( 1+{{x}^{2}} \right)}$$. **Note:** We can also do by another method by taking x as $$\tan \theta $$ and then replacing it in $${{\sin }^{-1}}\left( \dfrac{2x}{1+{{x}^{2}}} \right)$$ as $${{\sin }^{-1}}\left( \dfrac{2\tan \theta }{1+{{\tan }^{2}}\theta } \right)$$ and then using identity that $$\sin 2\theta =\dfrac{2\tan \theta }{1+{{\tan }^{2}}\theta }$$. So we can write it as $${{\sin }^{-1}}\left( \sin 2\theta \right)$$ or $$2\theta $$ which is $$2{{\tan }^{-1}}\left( x \right)$$.