Question

What is the derivative of $\arctan \left( \dfrac{x}{2} \right)$ ?

Answer 1

Hint: We know that ‘arctan’ is a term used for writing the inverse trigonometric relation. Thus, our problem is basically to find the derivative of ${{\tan }^{-1}}\left( \dfrac{x}{2} \right)$ . We will use the chain rule to find the derivative of ${{\tan }^{-1}}\left( \dfrac{x}{2} \right)$ with respect to $\left( \dfrac{x}{2} \right)$ and then multiply it with the derivative of $\left( \dfrac{x}{2} \right)$ with respect to $(x)$. This will give us the required solution.

Complete step-by-step solution:
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:
$\Rightarrow y={{\tan }^{-1}}(x)$
Then, we need to find the differential of ‘y’ with respect to ‘x’. This can be done as follows:
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left[ {{\tan }^{-1}}\left( \dfrac{x}{2} \right) \right]}{dx}$
On applying chain rule, we can simplify the above equation as follows:
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left[ {{\tan }^{-1}}\left( \dfrac{x}{2} \right) \right]}{d\left( \dfrac{x}{2} \right)}\times \dfrac{d\left( \dfrac{x}{2} \right)}{dx}$
Let us name the above equation as $(1)$, so we have:
$\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left[ {{\tan }^{-1}}\left( \dfrac{x}{2} \right) \right]}{d\left( \dfrac{x}{2} \right)}\times \dfrac{d\left( \dfrac{x}{2} \right)}{dx}$ ………$(1)$
Now, the formula for differential of inverse tangent is equal to:
$\Rightarrow \dfrac{d\left( {{\tan }^{-1}}\theta \right)}{d\theta }=\dfrac{1}{1+{{\theta }^{2}}}$
Using this formula in equation number $\left( 1 \right)$, we get:
$\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{1+{{\left( \dfrac{x}{2} \right)}^{2}}}\times \left( \dfrac{1}{2} \right)$
On further simplifying the above equation, we get:
$\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{1+\dfrac{{{x}^{2}}}{4}}\times \left( \dfrac{1}{2} \right) \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{4}{4+{{x}^{2}}}\times \left( \dfrac{1}{2} \right) \\
 & \therefore \dfrac{dy}{dx}=\dfrac{2}{4+{{x}^{2}}} \\
\end{align}$
Therefore, we can write the final required equation as:
$\Rightarrow \dfrac{d\left[ \arctan \left( \dfrac{x}{2} \right) \right]}{dx}=\dfrac{2}{4+{{x}^{2}}}$
Hence, the derivative of $\arctan \left( \dfrac{x}{2} \right)$ comes out to be $\dfrac{2}{4+{{x}^{2}}}$ .

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.