Question

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

Answer 1

Hint: To find the derivative of ${{\cot }^{-1}}\left( x \right)$ , we have to equate y to ${{\cot }^{-1}}\left( x \right)$ . From this, we will get $\cot y=x$ . We have to differentiate this equation with respect to x. Then, we have to use the trigonometric identities to solve further. Finally, we will have to use above substitutions.

Complete step by step solution:
We have to find the derivative of ${{\cot }^{-1}}\left( x \right)$ . Let us first equate y to ${{\cot }^{-1}}\left( x \right)$ .
$\Rightarrow y={{\cot }^{-1}}\left( x \right)...\left( i \right)$
We can write the above form as
$\Rightarrow \cot y=x...\left( ii \right)$
Let us differentiate the above equation with respect to x. We know that derivative of $\cot x$ is $-{{\csc }^{2}}x$ and $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ . Therefore, differentiation of the above equation gives
$\Rightarrow -{{\csc }^{2}}y\dfrac{dy}{dx}=1$
Let us take $-{{\csc }^{2}}y$ to the RHS.
$\Rightarrow \dfrac{dy}{dx}=-\dfrac{1}{{{\csc }^{2}}y}...\left( iii \right)$
We know that
\[{{\csc }^{2}}x-{{\cot }^{2}}x=1\]
We can rearrange the terms of this equation so that we can get the identity for ${{\csc }^{2}}x$ .
\[\Rightarrow {{\csc }^{2}}x=1+{{\cot }^{2}}x\]
We have to substitute the above formula in equation (iii).
$\Rightarrow \dfrac{dy}{dx}=-\dfrac{1}{1+{{\cot }^{2}}y}$
Let us substitute for $\cot y$ from equation (ii) in the above equation.
$\Rightarrow \dfrac{dy}{dx}=-\dfrac{1}{1+{{x}^{2}}}$
We can now substitute for y from equation (i) in the above equation.
$\Rightarrow \dfrac{d}{dx}{{\cot }^{-1}}x=-\dfrac{1}{1+{{x}^{2}}}$

Hence, the derivative of ${{\cot }^{-1}}\left( x \right)$ is $-\dfrac{1}{1+{{x}^{2}}}$

Note: Students must never miss out to substitute for $\cot y$ in the final step of differentiation. The final answer must be in terms of x. They must know the trigonometric identities and differentiation of basic functions. Students should never forget to write $\dfrac{dy}{dx}$ when differentiating $\cot y$ with respect to x. They may forget to put the negative sign in the derivative of $\cot y$ .