Question

What is the derivative of ${{\cos }^{-1}}\left( x \right)$ ?

Answer 1

Hint: We need to find the derivative of the function ${{\cos }^{-1}}\left( x \right)$ . We start to solve the problem by implicitly differentiating the given function. Then, we use the trigonometric identities and formulae to simplify the derivative to get the desired result.

Complete step by step solution:
We are given a function and need to find the derivative of it. We solve this question using implicit differentiation.
In implicit differentiation, we find the derivatives of the two variables by treating one variable as the function of the other.
Let us consider that the variable y is equal to the given function.
Writing the same, we get,
$\Rightarrow y={{\cos }^{-1}}x$
${{\cos }^{-1}}$ on shifting to another side of the equation becomes $\cos$
Shifting ${{\cos }^{-1}}$ to the other side of the equation, we get,
$\Rightarrow \cos y=x$
We need to differentiate the above equation implicitly with respect to x.
Differentiating the above equation, we get,
From the formulae of differentiation,
We know that the derivative of $\cos x$ is $-\sin x$ .
Similarly,
The derivative of the function $\cos y$ is given by $-\sin y\dfrac{dy}{dx}$ as the differentiation is with respect to variable x.
The derivative of x is 1.
Substituting the above values in the equation, we get,
$\Rightarrow -\sin y\dfrac{dy}{dx}=1$
Now,
From trigonometric identities,
We know that ${{\sin }^{2}}y+{{\cos }^{2}}y=1$
From the above, we know that
$\Rightarrow \cos y=x$
Squaring the above equation on both sides, we get,
$\Rightarrow {{\cos }^{2}}y={{x}^{2}}$
Substituting the value of ${{\cos }^{2}}y$ in the trigonometric identity, we get,
$\Rightarrow {{\sin }^{2}}y+{{x}^{2}}=1$
Shifting the value of ${{x}^{2}}$ to the other side of equation, we get,
$\Rightarrow {{\sin }^{2}}y=1-{{x}^{2}}$
$\therefore \sin y=\sqrt{1-{{x}^{2}}}$
Substituting the value of $\sin y$ in the equation $-\sin y\dfrac{dy}{dx}=1$ ,
$\Rightarrow -\sqrt{1-{{x}^{2}}}\dfrac{dy}{dx}=1$
Finding the value of $\dfrac{dy}{dx}$ , we get,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{-1}{\sqrt{1-{{x}^{2}}}}$
From the above, we know that $y={{\cos }^{-1}}x$
Substituting the same, we get,
$\therefore \dfrac{d}{dx}\left( {{\cos }^{-1}}x \right)=\dfrac{-1}{\sqrt{1-{{x}^{2}}}}$

Note: One must remember that the inverse of $\cos x$ is not the same as $\dfrac{1}{\cos x}$ . The given question can be also solved using the chain rule. The chain rule is used to find the derivatives of the composite functions.