Question

How do you find the derivative of $y=\arcsin \left( {{x}^{2}} \right)$?

Answer 1

In this question, we need to find the derivative of a function $y=\arcsin \left( {{x}^{2}} \right)$. For this we require $\dfrac{dy}{dx}$. We will consider ${{x}^{2}}$ as g(x) and the whole function as y = f(g(x)). To find the derivative with respect to x we will use chain rule by which $\dfrac{d}{dx}f\left( g\left( x \right) \right)=f'\left( g\left( x \right) \right)\cdot g'\left( x \right)$ where f'(g(x)) is the derivative of f(g(x)) with respect to g(x). We will use the general formula of derivative of arcsinx which is $\dfrac{d}{dx}\arcsin x=\dfrac{1}{\sqrt{1-{{x}^{2}}}}$ and also derivative of ${{x}^{n}}$ which is equal to $n{{x}^{n-1}}$.

Complete step by step answer:
Here we are given the function as $y=\arcsin \left( {{x}^{2}} \right)$. We need to find its derivative with respect to x. For this let us first suppose ${{x}^{2}}$ as g(x). Hence our function becomes $y=\arcsin \left( g\left( x \right) \right)$.
To apply the chain rule let us suppose arcsin(g(x)) as f(g(x)).
We know that according to chain rule $\dfrac{d}{dx}f\left( g\left( x \right) \right)=f'\left( g\left( x \right) \right)\cdot g'\left( x \right)$ where f'(g(x)) is the derivative of f(g(x)) with respect to g(x). Let us try to find $\dfrac{dy}{dx}$ if $y=\arcsin \left( g\left( x \right) \right)$.
We know that the derivative of arcsinx with respect to x is given by $\dfrac{1}{\sqrt{1-{{x}^{2}}}}$.
Hence derivative of arcsin(g(x)) with respect to g(x) will be $\dfrac{1}{\sqrt{1-{{\left( g\left( x \right) \right)}^{2}}}}$.
So we have $f'\left( g\left( x \right) \right)=\dfrac{1}{\sqrt{1-{{\left( g\left( x \right) \right)}^{2}}}}$.
Using chain rule we get $\dfrac{dy}{dx}=\dfrac{d\arcsin \left( g\left( x \right) \right)}{dx}=\dfrac{1}{\sqrt{1-{{\left( g\left( x \right) \right)}^{2}}}}\cdot g'\left( x \right)$.
We have g(x) as ${{x}^{2}}$.
We know that the derivative of ${{x}^{n}}$ is given by $n{{x}^{n-1}}$. So, the derivative of $g\left( x \right)={{x}^{2}}$ will be given by g'(x) = 2x. Putting in the value of g(x) and g'(x) we get $\dfrac{dy}{dx}=\dfrac{d}{dx}\arcsin {{\left( x \right)}^{2}}=\dfrac{1}{\sqrt{1-{{\left( {{\left( x \right)}^{2}} \right)}^{2}}}}\cdot 2x$.
Using ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$ in the denominator and arranging the terms we get $\dfrac{dy}{dx}=\dfrac{2x}{\sqrt{1-{{x}^{4}}}}$.
Hence the derivative of $y=\arcsin \left( {{x}^{2}} \right)$ is equal to $\dfrac{dy}{dx}=\dfrac{2x}{\sqrt{1-{{x}^{4}}}}$.

Note:
Students should note that arcsin and ${{\sin }^{-1}}x$ are the same. They should keep in mind the derivative of all inverse trigonometric functions. Note that x has been squared twice in the denominator so do not forget to square the ${{x}^{2}}$.