Question

What is the second derivative of inverse tangent?

Answer 1

Hint: To solve this question, we should know about the derivatives and the derivation formulas.This is because we can calculate the second derivative by the solving of the first derivative’s derivation again. The second derivative of any function is always the same if we calculate that by the formula or by that derivative again.

Complete step by step answer:
In order to solve this question, we will first start by considering $y=\arctan x$. It is the same as $y={{\tan }^{-1}}x$. Now let us move the inverse tangent to the left side of the equation. By doing so, it will give us,
$\tan y=x\ldots \ldots \ldots \left( 1 \right)$
Now what we want to do here is something called implicit differentiation. In this we want to take the derivatives of both sides of the equation.
$\dfrac{dy}{dx}{{\sec }^{2}}y=1$
Now, dividing both sides of the equation by ${{\sec }^{2}}y$, it will give us,
$\dfrac{dy}{dx}=\dfrac{1}{{{\sec }^{2}}\left( y \right)}$
We know that ${{\tan }^{2}}\theta +1={{\sec }^{2}}\theta $. So, by substituting this identity in our equation, it will give us,
$\dfrac{dy}{dx}=\dfrac{1}{{{\tan }^{2}}\left( y \right)+1}\ldots \ldots \ldots \left( 2 \right)$
Now, by the above defined equation number (1), we recall that $\tan y=x$. So, by substituting this into equation number (2), it will give us,
$\dfrac{dy}{dx}=\dfrac{1}{{{x}^{2}}+1}$
Now, the second derivative of $\arctan $ can be directly calculated by the Quotient Rule formula. The formula of the Quotient Rule is give as,
$f'\left( x \right)=\dfrac{g'\left( x \right)h\left( x \right)-h'\left( x \right)g\left( x \right)}{{{\left| h\left( x \right) \right|}^{2}}}$
The second derivative will be,
$\begin{align}
  & \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{\left( {{x}^{2}}+1 \right)\left( 0 \right)-\left( 1 \right)\left( 2x \right)}{{{\left( {{x}^{2}}+1 \right)}^{2}}} \\
 & \Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{-2x}{{{\left( {{x}^{2}}+1 \right)}^{2}}} \\
\end{align}$

Note: For all the inverse trigonometric functions first differentiate them and then all of them have general procedure. For example, when taking the derivative of $\arctan $, we move the inverse tangent to the right side (other side), do implicit differentiation and then use trigonometric identity.