Question

How do you find the derivative of the function $y=\arccos \left( \arcsin x \right)$?

Answer 1

Hint: In the above given question, we have a function which is a composite function which cannot be solved directly. Therefore, to solve this function we will use the chain rule which is written as $F'(x)=f'(g(x))g'(x)$and then simplify the expression to get the required solution.

Complete step by step solution:
We have the function given to us as:
$\Rightarrow y=\arccos \left( \arcsin x \right)$
We have to find the derivative of the given expression therefore; it can be written as:
$\Rightarrow y'=\dfrac{d}{dx}\arccos \left( \arcsin x \right)$
Now the expression in the form of a composite derivative therefore, we will use the chain rule which is: $F'(x)=f'(g(x))g'(x)$
In this case we have $g(x)=\arcsin x$.
Now we know that $\dfrac{d}{dx}\arccos \left( \dfrac{x}{a} \right)=\dfrac{-1}{\sqrt{{{a}^{2}}-{{x}^{2}}}}$
Now in this case we have no denominator therefore, $a=1$ ,on differentiating, we get:
$\Rightarrow y'=-\dfrac{1}{\sqrt{1-{{\arcsin }^{2}}x}}\dfrac{d}{dx}\arcsin x$
Now we know that $\dfrac{d}{dx}\arcsin \left( \dfrac{x}{a} \right)=\dfrac{1}{\sqrt{{{a}^{2}}-{{x}^{2}}}}$
Now in this case we have no denominator therefore, $a=1$ ,on differentiating, we get:
$\Rightarrow y'=-\dfrac{1}{\sqrt{1-{{\arcsin }^{2}}x}}\dfrac{1}{\sqrt{1-{{x}^{2}}}}$
On multiplying the terms, we get:
$\Rightarrow y'=-\dfrac{1}{\sqrt{1-{{x}^{2}}}\sqrt{1-{{\arcsin }^{2}}x}}$, which is the required solution.

Note: It is to be remembered that chain rule is used only when the expression is in the form of a composite function, which means it is in the form of $f(g(x))$.
It is to be remembered that integration and differentiation are inverse of each other. If the derivative of the term $X$ is $Y$, then inversely, the integration of $Y$ will be $X$.
We have used the $\arccos x$ and the $\arcsin x$ trigonometric functions over here, which are used to find the angle from the value of the trigonometric expression $\cos x$ and $\sin x$ respectively.
It is to be remembered that another way of writing $\arccos x$ and $\arcsin x$ function is by using the inverse notation which is written as ${{\sin }^{-1}}x$ and ${{\cos }^{-1}}x$ respectively.