How do you differentiate $y = {\cos ^{ - 1}}(1 - 2{x^2})$?

Hint: In the above question, we have a composite expression therefore, there is no direct formula to find its derivative thus, and we will use the chain rule to find the derivative of the given expression.

Complete step-by-step solution:
We have the given expression as:
$ \Rightarrow y = {\cos ^{ - 1}}(1 - 2{x^2})$
We have to find the derivative of the given expression therefore; it can be written as:
$ \Rightarrow y' = \dfrac{d}{{dx}}{\cos ^{ - 1}}(1 - 2{x^2})$
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 $g(x) = 1 - 2{x^2}$.
Now we know that $\dfrac{d}{{dx}}{\cos ^{ - 1}}\left( {\dfrac{x}{a}} \right) = \dfrac{{ - 1}}{{\sqrt {{a^2} - {x^2}} }}$
Now in this case we have no denominator therefore, $a = 1$, on differentiating, we get:
$ \Rightarrow y' = \dfrac{{ - 1}}{{\sqrt {1 - {{(1 - 2{x^2})}^2}} }}\dfrac{d}{{dx}}(1 - 2{x^2})$
Now we know that $\dfrac{d}{{dx}}k = 0$ and $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$ therefore, on differentiating, we get:
$ \Rightarrow y' = \dfrac{{ - 1}}{{\sqrt {1 - {{(1 - 2{x^2})}^2}} }} \times (2( - 2x))$
On simplifying, we get:
$ \Rightarrow y' = \dfrac{{4x}}{{\sqrt {1 - {{(1 - 2{x^2})}^2}} }}$
On expanding the square term in the denominator using the formula ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$, we get:
$ \Rightarrow y' = \dfrac{{4x}}{{\sqrt {1 - (1 - 4{x^2} + 4{x^4})} }}$
On opening the bracket, we get:
$ \Rightarrow y' = \dfrac{{4x}}{{\sqrt {1 - 1 + 4{x^2} - 4{x^4}} }}$
Now on simplifying, we get:
$ \Rightarrow y' = \dfrac{{4x}}{{\sqrt {4{x^2} - 4{x^4}} }}$
Now the term $4{x^2}$ is common in both the terms, we can take it out as common and write the expression as:
$ \Rightarrow y' = \dfrac{{4x}}{{\sqrt {4{x^2}(1 - {x^2})} }}$
Now since the $\sqrt {4{x^2}} = 2x$, we take the term out of the square root and write it as:
$ \Rightarrow y' = \dfrac{{4x}}{{2x\sqrt {1 - {x^2}} }}$
Now on simplifying, we get:

$ \Rightarrow y' = \dfrac{2}{{\sqrt {1 - {x^2}} }}$, 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 ${\cos ^{ - 1}}x$ trigonometric function over here, which is used to find the angle from the value of the trigonometric expression $\cos x$.