Question

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

Answer 1

Hint: The derivative of ${\sec }^2\left(x\right)$ can be found using the quotient and the chain rule in trigonometry. Firstly, We find the derivative of $\sec x$ by considering $u=\sec \left(x\right)$ . Then, we find the derivative of $u^2$ to get the required result.

Complete step-by-step solution:
We are given a function and need to find the derivative of it. We solve this question using the quotient and the chain rule in differentiation.
The chain rule is used to find the derivatives of the composite functions.
Let us consider,
$\Rightarrow u=\sec x$
Differentiating the above equation on both sides with respect to $x$ , we get,
$\Rightarrow {\displaystyle \frac{du}{dx}}={\displaystyle \frac{d}{dx}}(\sec x)$
From trigonometry,
We know that secant function is the inverse of the cosine function.
$\Rightarrow \sec x={\displaystyle \frac{1}{\cos x}}$
Substituting the same, we get,
$\Rightarrow {\displaystyle \frac{du}{dx}}={\displaystyle \frac{d}{dx}}\left({\displaystyle \frac{1}{\cos x}}\right)$
The Quotient rule of differentiation is given as follows,
$\Rightarrow {\displaystyle \frac{d}{dx}}\left({\displaystyle \frac{m}{n}}\right)={\displaystyle \frac{(n{\displaystyle \frac{dm}{dx}}-m{\displaystyle \frac{dn}{dx}})}{n^2}}$
Applying the quotient rule of differentiation to the above equation, we get,
$\Rightarrow {\displaystyle \frac{du}{dx}}=$${\displaystyle \frac{(\cos x{\displaystyle \frac{d\left(1\right)}{dx}}-1{\displaystyle \frac{d(\cos x)}{dx}})}{{\cos }^{2}x}}$
The derivative of any constant function c is always zero expressed as follows,
$\Rightarrow {\displaystyle \frac{d\left(c\right)}{dx}}=0$
The derivative of the cosine function is negative of sine function expressed as follows,
$\Rightarrow {\displaystyle \frac{d}{dx}}(\cos x)=-\sin x$
Substituting the values in the above expression,
$\Rightarrow {\displaystyle \frac{du}{dx}}={\displaystyle \frac{(0-1(-\sin x))}{{\cos }^{2}x}}$
Let us evaluate it further.
$\Rightarrow {\displaystyle \frac{du}{dx}}={\displaystyle \frac{\sin x}{{\cos }^{2}x}}$
Splitting the denominator, we get,
$\Rightarrow {\displaystyle \frac{du}{dx}}={\displaystyle \frac{\sin x}{\cos x}}\times {\displaystyle \frac{1}{\cos x}}$
$\Rightarrow {\displaystyle \frac{du}{dx}}=\tan x\sec x$
Substituting the value of $u$ in the above equation,
$\therefore {\displaystyle \frac{d(\sec x)}{dx}}=\tan x\sec x$
Now,
Let us consider a variable such that,
$\Rightarrow v=u^2$
Differentiating the equation on both sides, we get,
$\Rightarrow {\displaystyle \frac{dv}{dx}}={\displaystyle \frac{d\left(u^2\right)}{dx}}$
$\Rightarrow {\displaystyle \frac{dv}{dx}}=2u{\displaystyle \frac{du}{dx}}$
Substituting the value of ${\displaystyle \frac{du}{dx}}$ from the above, we get,
 $\Rightarrow {\displaystyle \frac{dv}{dx}}=2u\times \tan x\sec x$
We know that $u=\sec x$ . Substituting the value of $u$ , we get,
$\Rightarrow {\displaystyle \frac{dv}{dx}}=2\times \sec x\times \tan x\sec x$
$\Rightarrow {\displaystyle \frac{dv}{dx}}=2{\sec }^{2}x\tan x$
Substituting the value of $v$ ,
$\Rightarrow {\displaystyle \frac{d\left(u^2\right)}{dx}}=2{\sec }^{2}x\tan x$
$\therefore {\displaystyle \frac{d\left({\sec }^{2}x\right)}{dx}}=2{\sec }^{2}x\tan x$

Note: We must remember that the derivative of any constant function is zero. One of the most common mistakes in derivatives is choosing the right rule of differentiation.
1. If two functions are multiplying each other, we must apply the product rule.
2. If two functions are dividing each other, we must apply the quotient rule.
3. If the given function is composite, we must apply the chain rule.