Question

How do you find the derivative of ${\cos ^{ - 1}}\left( {\dfrac{x}{2}} \right)$?

Answer 1

In this question we have to find the derivative of the given inverse function, first assume the given function as a variable, and then apply the cos to both sides, and then derive both sides of the equation we will get an expression in terms of $\sin \left( {{{\cos }^{ - 1}}\dfrac{x}{2}} \right)$, now using the trigonometric identity ${\sin ^2}x + {\cos ^2}x = 1$, let’s assume here as, $x = {\cos ^{ - 1}}\dfrac{x}{2}$, then simplify the expression and to get the value of $\sin \left( {{{\cos }^{ - 1}}\dfrac{x}{2}} \right)$and then by substituting the value in the first expression and then further simplification we will get the required result.

Complete step by step solution:
Given function is ${\cos ^{ - 1}}\left( {\dfrac{x}{2}} \right)$,
Now let’s assume the given function as $y$,
$\Rightarrow y = {\cos ^{ - 1}}\dfrac{x}{2}$,
Now apply cos on both sides we get,
$\Rightarrow \cos y = \cos \left( {{{\cos }^{ - 1}}\dfrac{x}{2}} \right)$,
Now simplifying we get,
$\Rightarrow \cos y = \dfrac{x}{2}$,
Now differentiate in both sides of the equation we get,
$\Rightarrow \dfrac{d}{{dx}}\cos y = \dfrac{d}{{dx}}\dfrac{x}{2}$,
Now simplifying we get,
$\Rightarrow - \sin y\dfrac{{dy}}{{dx}} = \dfrac{1}{2}$ ,
Now taking sin y to the right hand side we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{1}{2}\dfrac{1}{{\sin y}}$,
Now we know that $y = {\cos ^{ - 1}}\dfrac{x}{2}$, then the equation becomes,
$\Rightarrow \dfrac{d}{{dx}}{\cos ^{ - 1}}\dfrac{x}{2} = \dfrac{{ - 1}}{{2\sin \left( {{{\cos }^{ - 1}}\left( {\dfrac{x}{2}} \right)} \right)}}$,-----(1),
Now from the trigonometric identity ${\sin ^2}x + {\cos ^2}x = 1$, let’s assume here as, $x = {\cos ^{ - 1}}\dfrac{x}{2}$, then the identity becomes,
$\Rightarrow {\sin ^2}\left( {{{\cos }^{ - 1}}\dfrac{x}{2}} \right) + {\cos ^2}\left( {{{\cos }^{ - 1}}\left( {\dfrac{x}{2}} \right)} \right) = 1$,
Now rewrite the expression as,
$\Rightarrow {\sin ^2}\left( {{{\cos }^{ - 1}}\dfrac{x}{2}} \right) + {\left( {\cos \left( {{{\cos }^{ - 1}}\left( {\dfrac{x}{2}} \right)} \right)} \right)^2} = 1$,
Now we know that $\cos \left( {{{\cos }^{ - 1}}x} \right) = x$ the identity becomes,
$\Rightarrow {\sin ^2}\left( {{{\cos }^{ - 1}}\dfrac{x}{2}} \right) + \dfrac{{{x^2}}}{4} = 1$,
Now simplifying we get,
$\Rightarrow {\sin ^2}\left( {{{\cos }^{ - 1}}\dfrac{x}{2}} \right) = 1 - \dfrac{{{x^2}}}{4}$,
Now taking out the square root we get,
$\sin \left( {{{\cos }^{ - 1}}\dfrac{x}{2}} \right) = \sqrt {1 - \dfrac{{{x^2}}}{4}}$,
Now simplifying we get,
$\sin \left( {{{\cos }^{ - 1}}\dfrac{x}{2}} \right) = \sqrt {\dfrac{{4 - {x^2}}}{4}}$,
Now again simplifying we get,
$\Rightarrow \sin \left( {{{\cos }^{ - 1}}\dfrac{x}{2}} \right) = \dfrac{1}{2}\sqrt {4 - {x^2}}$,
Now substituting the value in (1) we get,
$\dfrac{d}{{dx}}{\cos ^{ - 1}}\left( {\dfrac{x}{2}} \right) = \dfrac{{ - 1}}{{2\dfrac{1}{2}\sqrt {4 - {x^2}} }}$,
Now simplifying we get,
$\dfrac{d}{{dx}}{\cos ^{ - 1}}\left( {\dfrac{x}{2}} \right) = \dfrac{{ - 1}}{{\sqrt {4 - {x^2}} }}$.
So, the derivative of the given function is $\dfrac{{ - 1}}{{\sqrt {4 - {x^2}} }}$.

Final Answer:
$\therefore$ The derivative of the given function ${\cos ^{ - 1}}\left( {\dfrac{x}{2}} \right)$ will be equal to $\dfrac{{ - 1}}{{\sqrt {4 - {x^2}} }}$.

Note:
Another method of solving the question is by directly using the derivative of inverse trigonometric functions the formula which is given by $\dfrac{d}{{dx}}{\cos ^{ - 1}}x = \dfrac{{ - 1}}{{\sqrt {1 - {x^2}} }}$,
Now given function is ${\cos ^{ - 1}}\left( {\dfrac{x}{2}} \right)$,
Now here $x = \dfrac{x}{2}$, by substitute the value in the formula we get,
$\Rightarrow \dfrac{d}{{dx}}{\cos ^{ - 1}}\dfrac{x}{2} = \dfrac{{ - 1}}{{\sqrt {1 - \dfrac{{{x^2}}}{4}} }}\dfrac{d}{{dx}}\dfrac{x}{2}$,
Now simplifying we get,
$\Rightarrow \dfrac{d}{{dx}}{\cos ^{ - 1}}\dfrac{x}{2} = \dfrac{{ - 1}}{{\sqrt {\dfrac{{4 - {x^2}}}{4}} }}\dfrac{1}{2}$,
Now simplifying we get,
$\Rightarrow \dfrac{d}{{dx}}{\cos ^{ - 1}}\dfrac{x}{2} = \dfrac{{ - 2}}{{\sqrt {4 - {x^2}} }}\dfrac{1}{2}$,
Now further simplifying we get,
$\Rightarrow \dfrac{d}{{dx}}{\cos ^{ - 1}}\dfrac{x}{2} = \dfrac{{ - 1}}{{\sqrt {4 - {x^2}} }}$,
So, we got the same derivative value i.e., $\dfrac{{ - 1}}{{\sqrt {4 - {x^2}} }}$.