Question

How do you differentiate $ \text{arc tan}\left( {{x}^{2}} \right) $ ?

Answer 1

Hint: In this question, we need to find the derivative of the function $ \text{arc tan}\left( {{x}^{2}} \right) $ . For this, we will consider $ {{x}^{2}} $ as g(x) and the whole function as f(g(x)). After that, we will apply chain rule on f(g(x)) to find the derivative of f(g(x)). By chain rule, $ \dfrac{d}{dx}f\left( g\left( x \right) \right)=f'\left( g\left( x \right) \right)\cdot g'\left( x \right) $ . Also we will use the general formula of derivative of arctanx i.e. $ \dfrac{d}{dx}\text{arc tanx}=\dfrac{1}{1+{{x}^{2}}} $ .

Complete step by step answer:
Here we are given the function as $ \text{arc tan}\left( {{x}^{2}} \right) $ .
Let us suppose that $ {{x}^{2}}=g\left( x \right) $ . So our function reduces to arctan(g(x)).
Now let us suppose the whole function is f(g(x)). Therefore we have $ f\left( g\left( x \right) \right)=\text{arc tan}\left( g\left( x \right) \right) $ .
We need to find the derivative of a given function with respect to x.
Therefore, we will use chain rule for that. But we know, for a composition f(g(x)), derivative with respect to x is given as, $ \dfrac{d}{dx}f\left( g\left( x \right) \right)=f'\left( g\left( x \right) \right)\cdot g'\left( x \right) $ .
Now we know that, derivative of arctanx with respect to x is given by $ \dfrac{1}{1+{{x}^{2}}} $ . So, derivative of arctan(g(x)) will be given by $ \dfrac{1}{1+{{\left( g\left( x \right) \right)}^{2}}}\cdot g'\left( x \right) $ . (By chain rule). Therefore we have,
 $ \dfrac{d}{dx}f\left( g\left( x \right) \right)=\dfrac{1}{1+{{\left( g\left( x \right) \right)}^{2}}}\cdot g'\left( x \right) $ .
Putting in the value of g(x) we have, $ \Rightarrow \dfrac{1}{1+{{\left( {{x}^{2}} \right)}^{2}}}\dfrac{d}{dx}\left( {{x}^{2}} \right) $ .
Now we know that, the derivative of $ {{x}^{n}} $ is given as $ \dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}} $ . Therefore, here we have n = 2 so we get,
 $ \dfrac{d}{dx}\left( {{x}^{2}} \right)=2{{x}^{2-1}}=2x $ .
Therefore, $ \dfrac{d}{dx}\left( \text{arc tan}\left( {{x}^{2}} \right) \right)=\dfrac{1}{1+{{\left( {{x}^{2}} \right)}^{2}}}\cdot 2x $ .
Using the property of exponents that $ {{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}} $ we have, \[\dfrac{d}{dx}\left( \text{arc tan}{{x}^{2}} \right)=\dfrac{1}{1+{{x}^{4}}}\cdot 2x\].
Simplifying we get \[\dfrac{d}{dx}\left( \text{arc tan}{{x}^{2}} \right)=\dfrac{2x}{1+{{x}^{4}}}\].
Therefore the required derivative of \[\text{arc tan}{{x}^{2}}\] is equal to \[\dfrac{2x}{1+{{x}^{4}}}\].

Note:
 Students should note that arctanx and $ {{\tan }^{-1}}x $ are the same. Students should keep in mind the general formula of the derivative of an inverse trigonometric function. To avoid confusion, just try to use $ {{x}^{2}} $ instead of x in the formula and then multiply the answer by the derivative of $ {{x}^{2}} $ to get the final answer. Note that, x has been squared twice so do not forget to take the power of x as 4 in the denominator.