Question

What is the derivative of, $\left[ \arctan \left( x \right)+\arctan \left( \dfrac{1}{x} \right) \right]$ ?

Answer 1

Hint: The first term in our expression is easily differentiable. Then, our problem is basically to find the derivative of $\left[ \arctan \left( \dfrac{1}{x} \right) \right]$. For the second term, we will use the chain rule to find the derivative of $\left[ \arctan \left( \dfrac{1}{x} \right) \right]$ with respect to $\left( \dfrac{1}{x} \right)$ and then multiply it with the derivative of $\left( \dfrac{1}{x} \right)$ with respect to $(x)$. This will give us the required solution.

Complete step by step answer:
We know that ‘arctan’ is a term used for writing the inverse trigonometric relation. So, we can write our expression as: $\left[ ta{{n}^{-1}}\left( x \right)+{{\tan }^{-1}}\left( \dfrac{1}{x} \right) \right]$.
Now, let us first assign some terms that we are going to use later in our problem.
Let the given term on which we need to operate a differential be given by ‘y’ . Here, ‘y’ is given to us as:
$\Rightarrow y=\left[ ta{{n}^{-1}}\left( x \right)+{{\tan }^{-1}}\left( \dfrac{1}{x} \right) \right]$
Then, we need to find the differential of ‘y’ with respect to ‘x’. This can be done as follows:
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left[ ta{{n}^{-1}}\left( x \right)+{{\tan }^{-1}}\left( \dfrac{1}{x} \right) \right]}{dx}$
Applying the chain rule of differentiation on the second term, our expression can be written as:
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left[ {{\tan }^{-1}}x \right]}{dx}+\dfrac{d\left[ {{\tan }^{-1}}\left( \dfrac{1}{x} \right) \right]}{d\left( \dfrac{1}{x} \right)}\times \dfrac{d\left( \dfrac{1}{x} \right)}{dx}$
Using the formula for differential of inverse tangent that is equal to:
$\Rightarrow \dfrac{d\left( {{\tan }^{-1}}\theta \right)}{d\theta }=\dfrac{1}{1+{{\theta }^{2}}}$
We get:
$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{1+{{x}^{2}}}+\dfrac{1}{1+{{\left( \dfrac{1}{x} \right)}^{2}}}\times -\dfrac{1}{{{x}^{2}}}$ $\left[ \because \dfrac{d\left( \dfrac{1}{x} \right)}{dx}=-\dfrac{1}{{{x}^{2}}} \right]$
On further simplification, our expression becomes:
$\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{1+{{x}^{2}}}+\dfrac{{{x}^{2}}}{1+{{x}^{2}}}\times \left( -\dfrac{1}{{{x}^{2}}} \right) \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{1+{{x}^{2}}}-\dfrac{1}{1+{{x}^{2}}} \\
 & \therefore \dfrac{dy}{dx}=0 \\
\end{align}$

Hence, the derivative of $\left[ \arctan \left( x \right)+\arctan \left( \dfrac{1}{x} \right) \right]$ comes out to be Zero.

Note: It is very important to know the meaning of terms like ‘arctan’ or ‘arcsine’ or ‘arccosine’, etc. as these are some very common terms. Also, one should know the differential result of these trigonometric quantities as they are some very important results. Although they can be very easily derived, it is recommended to remember them thoroughly as it will save time spent in extra calculation.