Question

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

Answer 1

Hint: In this question we will find the derivative of the given trigonometric function by using the chain rule since it is in the form of a composite function. It has two functions in it which is the $\arcsin x$function, which is also called the inverse sine function. And we have the square function. We will use the formula of derivative that $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$and $\dfrac{d}{dx}(\arcsin x)=\dfrac{1}{\sqrt{1-{{x}^{2}}}}$to substitute in the expression and simplify to get the required solution.

Complete step-by-step solution:
We have the expression given to us as ${{\left( \arcsin x \right)}^{2}}$
Since we have to take the derivative of the term, we can write it as:
$\Rightarrow \dfrac{d}{dx}{{\left( \arcsin x \right)}^{2}}$
Now since there is no direct formula for calculating the derivative of the given expression, we will use the chain rule. The formula for the chain rule is: $F'(x)=f'(g(x))g'(x)$.
Now in this question we have a composite function in the form of $f(g(x))$where $f(x)={{x}^{2}}$ and $g(x)=\arcsin x$.
Now we know that $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ and since we have to use the chain rule on the expression, we can write it as:
$\Rightarrow 2(\arcsin x)\times \dfrac{d}{dx}(\arcsin x)$
Now we know that $\dfrac{d}{dx}(\arcsin x)=\dfrac{1}{\sqrt{1-{{x}^{2}}}}$ therefore, on using the formula and substituting, we get:
$\Rightarrow 2(\arcsin x)\times \dfrac{1}{\sqrt{1-{{x}^{2}}}}$
On simplifying the terms by multiplying, we get:
$\Rightarrow \dfrac{2(\arcsin x)}{\sqrt{1-{{x}^{2}}}}$, which is the required solution.

Note: In this question we have the inverse trigonometric function $\arcsin x$ which is called the inverse sine function or the sine inverse function. It is also represented as ${{\sin }^{-1}}x$.
The chain rule is to be used when there are multiple functions in the expression. In this question there are $2$ functions, there can be more than $2$ functions in an expression too.
The inverse of the derivative is the integration and vice versa. If the derivative of a term $a$ is $b$, then the integration of the term $b$ will be $a$.