Question

How do you differentiate $f(x) = {\sec ^2}x - {\tan ^2}x$ ?

Answer 1

Hint: In this question, we need to differentiate the given function. Firstly we make use of difference rule of differentiation which is given by $\dfrac{d}{{dx}}(g(x) - h(x)) = \dfrac{d}{{dx}}g(x) - \dfrac{d}{{dx}}h(x)$. Now we differentiate the function $g(x)$ and $h(x)$. We also make use of chain rule which is given by $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}$. Then we simplify it and obtain the required derivative of the function.

Complete step-by-step answer:
Given a function of the form $f(x) = {\sec ^2}x - {\tan ^2}x$
We are asked to find the derivative of the above function.
i.e. we need to differentiate it with respect to x.
We know that if $g(x)$, $h(x)$ are two real valued functions, then the differentiation of their difference is given by the difference rule as,
$\dfrac{d}{{dx}}(g(x) - h(x)) = \dfrac{d}{{dx}}g(x) - \dfrac{d}{{dx}}h(x)$
Note that here $g(x) = {\sec ^2}x$ and $h(x) = {\tan ^2}x$.
So we have, $\dfrac{d}{{dx}}({\sec ^2}x - {\tan ^2}x) = \dfrac{d}{{dx}}{\sec ^2}x - \dfrac{d}{{dx}}{\tan ^2}x$ …… (1)
Now to differentiate the functions in the R.H.S. with respect to x, we make use of chain rule.
The chain rule is given by $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}$ …… (2)
Now we first find $\dfrac{d}{{dx}}{\sec ^2}x$.
Here we take $y = {\sec ^2}x$ and $u = \sec x$
So substituting this, we get,
$ \Rightarrow \dfrac{d}{{dx}}{\sec ^2}x = \dfrac{d}{{d(\sec x)}}({\sec ^2}x) \times \dfrac{d}{{dx}}(\sec x)$
We can write ${\sec ^2}x = {(\sec x)^2}$
Hence we get,
$ \Rightarrow \dfrac{d}{{dx}}{\sec ^2}x = \dfrac{d}{{d(\sec x)}}{(\sec x)^2} \times \dfrac{d}{{dx}}(\sec x)$
We know that $\dfrac{d}{{dt}}{t^n} = n{t^{n - 1}}$ and $\dfrac{d}{{dx}}(\sec x) = \sec x\tan x$
Here we have $t = \sec x$ and $n = 2$. Hence we get,
$ \Rightarrow \dfrac{d}{{dx}}{\sec ^2}x = 2{(\sec x)^{2 - 1}} \times \sec x\tan x$
$ \Rightarrow \dfrac{d}{{dx}}{\sec ^2}x = 2{(\sec x)^1} \times \sec x\tan x$
$ \Rightarrow \dfrac{d}{{dx}}{\sec ^2}x = 2{\sec ^2}x\tan x$ …… (3)
Now we find $\dfrac{d}{{dx}}{\tan ^2}x$ using the chain rule given in the equation (2).
Here we take $y = {\tan ^2}x$ and $u = \tan x$
So substituting this, we get,
$ \Rightarrow \dfrac{d}{{dx}}{\tan ^2}x = \dfrac{d}{{d(\tan x)}}({\tan ^2}x) \times \dfrac{d}{{dx}}(\tan x)$
We can write ${\tan ^2}x = {(\tan x)^2}$
Hence we get,
$ \Rightarrow \dfrac{d}{{dx}}{\tan ^2}x = \dfrac{d}{{d(\tan x)}}{(\tan x)^2} \times \dfrac{d}{{dx}}(\tan x)$
We know that $\dfrac{d}{{dt}}{t^n} = n{t^{n - 1}}$ and $\dfrac{d}{{dx}}(\tan x) = {\sec ^2}x$
Here we have $t = \tan x$ and $n = 2$. Hence we get,
$ \Rightarrow \dfrac{d}{{dx}}{\tan ^2}x = 2{(\tan x)^{2 - 1}} \times {\sec ^2}x$
$ \Rightarrow \dfrac{d}{{dx}}{\tan ^2}x = 2{(\tan x)^1} \times {\sec ^2}x$
$ \Rightarrow \dfrac{d}{{dx}}{\tan ^2}x = 2{\sec ^2}x\tan x$ …… (4)
Now substituting equations (3) and (4) in the equation (1), we get,
$\dfrac{d}{{dx}}({\sec ^2}x - {\tan ^2}x) = 2{\sec ^2}x\tan x - 2{\sec ^2}x\tan x$
$ \Rightarrow \dfrac{d}{{dx}}({\sec ^2}x - {\tan ^2}x) = 0$
Hence the differentiation of the function $f(x) = {\sec ^2}x - {\tan ^2}x$ is equal to $0$.
i.e. $\dfrac{d}{{dx}}({\sec ^2}x - {\tan ^2}x) = 0$.

Note:
The process to find the derivative of the function is called differentiation. In differentiation, there is an instantaneous rate of change of the functions based on the variable.
Students must remember some formulas of differentiation to find the derivative of a function.
Some of the differentiation formulas are as follows.
(1) $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$
(2) $\dfrac{d}{{dx}}cu = c\dfrac{d}{{dx}}u$ where, c is a constant.
(3) $\dfrac{d}{{dx}}u \pm v = \dfrac{d}{{dx}}u \pm \dfrac{d}{{dx}}v$
(4) $\dfrac{d}{{dx}}(\sec x) = \sec x\tan x$
(5) $\dfrac{d}{{dx}}(\tan x) = {\sec ^2}x$