Solveeit Logo

Question

Question: How do you take the derivative of \({{\tan }^{-1}}\left( {{x}^{2}} \right)\) ?...

How do you take the derivative of tan1(x2){{\tan }^{-1}}\left( {{x}^{2}} \right) ?

Explanation

Solution

Derivative is the rate of change of a function. To find the derivative of a function, we have to differentiate it with respect to xx . Finding out the derivative of a function implies that we are finding out the slope of the function. Here we have to use the Extension Chain Rule to solve this. We use this rule when the function , let’s say yy, is a function of uu and uu is a function of vv and vv is a function of xx. It states the following : dydx=dydu×dudv×dvdx\dfrac{dy}{dx}=\dfrac{dy}{du}\times \dfrac{du}{dv}\times \dfrac{dv}{dx} . So when we differentiate a function like that, we have to use this formula.

Complete step by step answer:
Here y=tan1(x2)y={{\tan }^{-1}}\left( {{x}^{2}} \right) , it is a function of tan1{{\tan }^{-1}} and tan1{{\tan }^{-1}} is a function of x2{{x}^{2}}.So we have to apply the Extension Chain rule in order to find the derivative of the given function. Upon comparing , uu is tan1x{{\tan }^{-1}}x and vv is x2{{x}^{2}} .
We already know that the derivative of tan1x{{\tan }^{-1}}x is
\Rightarrow d(tan1x)dx=11+x2\dfrac{d\left( {{\tan }^{-1}}x \right)}{dx}=\dfrac{1}{1+{{x}^{2}}} ( This is the derivative of tan1x{{\tan }^{-1}}x.)
d(xn)dx=nxn1\Rightarrow \dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}} which is the derivative of xn{{x}^{n}}
Here n=2n=2 , so d(x2)dx=2x\dfrac{d\left( {{x}^{2}} \right)}{dx}=2x ……. eqn(1)eqn\left( 1 \right)
Upon differentiating yywith respect to xx, we get the following :
dydx=d(tan1x2)dx×d(x2)dx\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( {{\tan }^{-1}}{{x}^{2}} \right)}{dx}\times \dfrac{d{{\left( {{x}^{2}} \right)}^{{}}}}{dx}
From eqn(1)eqn\left( 1 \right) :
dydx=11+x4×2x\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{1+{{x}^{4}}}\times 2x.
And we can further write it as
dydx=2x1+x4\Rightarrow \dfrac{dy}{dx}=\dfrac{2x}{1+{{x}^{4}}} .
dydx\dfrac{dy}{dx} here is nothing but the slope . So in other words the slope of the function tan1(x2)=2x1+x4{{\tan }^{-1}}\left( {{x}^{2}} \right)=\dfrac{2x}{1+{{x}^{4}}}

\therefore Hence , the derivative of tan1(x2){{\tan }^{-1}}\left( {{x}^{2}} \right) = 2x1+x4\dfrac{2x}{1+{{x}^{4}}}.

Note: In this question we have to be careful about a lot of things. First we should remember the derivatives of all function such as sinx,cos,tanx,xn,logx,ax,sin1x,cos1x,tan1x\sin x,\cos ,\tan x,{{x}^{n}},\log x,{{a}^{x}},{{\sin }^{-1}}x,{{\cos }^{-1}}x,{{\tan }^{-1}}x and all other related functions so that we can solve it easily. Secondly we should be careful while applying the Extension Chain Rule. If we get confused in applying, then the whole derivative may be wrong. It is also advisable to remember all the other standard rules of differentiation.