What is the derivative of $\sec x$?

Hint: In this question we have been asked to find the derivative of the given trigonometric function $\sec x$. We will first rewrite the expression in the form of $\cos x$ and then we will use the formula of the derivative of the term in the form of $\dfrac{u}{v}$. We will use the formula $\dfrac{d}{dx}\dfrac{u}{v}=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$ and simplify the terms to get the required solution.

Complete step-by-step solution:
We have the term given to us as:
$\Rightarrow \sec x$
Since we have to find the derivative of the term, it can be written as:
$\Rightarrow \dfrac{d}{dx}\sec x$
Now we know that $\sec x=\dfrac{1}{\cos x}$ therefore, on substituting, we get:
$\Rightarrow \dfrac{d}{dx}\dfrac{1}{\cos x}$
We can see that the expression is in the form of the derivative of $\dfrac{u}{v}$.
On using the formula $\dfrac{d}{dx}\dfrac{u}{v}=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$ on the expression, we get:
$\Rightarrow \dfrac{\cos x\dfrac{d}{dx}1-1\dfrac{d}{dx}\cos x}{{{\cos }^{2}}x}$
Now we know that $\dfrac{d}{dx}k=0$, where $k$ is any constant value and $\dfrac{d}{dx}\cos x=-\sin x$ therefore on substituting them in the expression, we get:
$\Rightarrow \dfrac{\cos x\left( 0 \right)-1\left( -\sin x \right)}{{{\cos }^{2}}x}$
On simplifying the terms, we get:
$\Rightarrow \dfrac{\sin x}{{{\cos }^{2}}x}$
Now the denominator can be split up and written as:
$\Rightarrow \dfrac{\sin x}{\cos x\times \cos x}$
Now we know that $\dfrac{\sin x}{\cos x}=\tan x$, therefore on substituting, we get:
$\Rightarrow \dfrac{\tan x}{\cos x}$
Now we know that $\dfrac{1}{\cos x}=\sec x$ therefore, on substituting, we get:
$\Rightarrow \sec x\tan x$, which is the required derivative.
Therefore, we can write:
$\Rightarrow \dfrac{d}{dx}\sec x=\sec x\tan x$

Note: It is to be remembered that the function we used to solve the expression is called as the quotient rule. There also exists another rule which is known as the product rule which deals with expressions in the form of $uv$ and has formula $\dfrac{d}{dx}uv=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$. It is to be noted that the terms $u$ and $v$ are also written as $f\left( x \right)$ and $g\left( x \right)$ in some solutions.