Question

If $y = {\tan ^{ - 1}}x$, find$\dfrac{{{d^2}y}}{{d{x^2}}}$ in terms of y alone.

Answer 1

In this question, we are given a simple equation. We have to differentiate the equation twice in order to find the double derivative. However, the answer should be in terms of y only. For this, convert the equation in terms of y in the beginning only rather than changing it at the end as this will make it much easier to solve. After doing so, simply differentiate it twice with respect to y in order to find the required answer.

Complete step-by-step answer:
We are given $y = {\tan ^{ - 1}}x$
Since the question asks for the answer in terms of y only, we will shift y to the other side which will convert our inverse trigonometric ratio to trigonometric ratio.
$\Rightarrow y = {\tan ^{ - 1}}x$
$\Rightarrow x = \tan y$
Now, differentiate the equation with respect to y,
$\Rightarrow \dfrac{{dx}}{{dy}} = {\sec ^2}y$
Next, we will reciprocate both the sides.
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{{{\sec }^2}y}}$
$\Rightarrow \dfrac{{dy}}{{dx}} = {\cos ^2}y$ …. ($\cos y = \dfrac{1}{{\sec y}}$)
Now, differentiate again but this time, with respect to x.
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = - 2{\cos ^2}y\sin y\dfrac{{dy}}{{dx}}$
We know that $\dfrac{{dy}}{{dx}} = {\cos ^2}y$. Putting this in the above equation,
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = - 2{\cos ^3}y\sin y$

Therefore $- 2{\cos ^3}y\sin y$ is the required answer.

Note: This question can also be solved by the following method. However, it is only for those students who find it difficult to deal with inverse trigonometry.
$\Rightarrow y = {\tan ^{ - 1}}x$ (given)
Take tan both the sides.
$\Rightarrow \tan y = \tan ({\tan ^{ - 1}}x)$
We know that $\tan ({\tan ^{ - 1}}x) = x$. Using this in the above equation,
$\Rightarrow \tan y = x$
After this step, this equation can be simply differentiated with respect to y and all the subsequent steps can be followed in the similar way.
Differentiating the equation with respect to y,
$\Rightarrow \dfrac{{dx}}{{dy}} = {\sec ^2}y$
Next, we will reciprocate both the sides.
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{{{\sec }^2}y}}$
$\Rightarrow \dfrac{{dy}}{{dx}} = {\cos ^2}y$ …. ($\cos y = \dfrac{1}{{\sec y}}$)
Now, differentiate again but this time, with respect to x.
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = - 2{\cos ^2}y\sin y\dfrac{{dy}}{{dx}}$
We know that $\dfrac{{dy}}{{dx}} = {\cos ^2}y$. Putting this in the above equation,
$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = - 2{\cos ^3}y\sin y$
The students can prefer method 1 or 2 if he/she is comfortable with inverse trigonometry. Otherwise, he/she can use method 2 as it is more elaborate and deals with basic rules of inverse trigonometry.