Question

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

Answer 1

Hint: 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.
\[\begin{align}
  & \dfrac{d}{dx}\left( {{\sin }^{-1}}x \right)=\dfrac{1}{\sqrt{1-{{x}^{2}}}} \\
 & \dfrac{d}{dx}\left( \dfrac{2x}{1+{{x}^{2}}} \right)=2x\left( \dfrac{d}{dx}\left( \dfrac{1}{1+{{x}^{2}}} \right) \right)+\dfrac{1}{1+{{x}^{2}}}\left( \dfrac{d}{dx}\left( 2x \right) \right) \\
\end{align}\]
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 \[2{{\tan }^{-1}}\left( x \right)\].