What is the derivative of $y=\tan \left( x \right)$?

Hint: In this question we have been asked to find the derivative of the given trigonometric function $\tan \left( x \right)$. We will first rewrite the expression in the form of the quotient of $\sin \left( x \right)$ and $\cos \left( x \right)$,then we will use the formula of 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:
$\Rightarrow y=\tan \left( x \right)$
Since we have to find the derivative of the term, it can be written as:
$\Rightarrow y'=\dfrac{d}{dx}\tan \left( x \right)$
Now we know that $\tan \left( x \right)=\dfrac{\sin \left( x \right)}{\cos \left( x \right)}$ therefore, on substituting, we get:
$\Rightarrow y'=\dfrac{d}{dx}\dfrac{\sin \left( x \right)}{\cos \left( x \right)}$
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:
$\Rightarrow y'=\dfrac{\cos \left( x \right)\dfrac{d}{dx}\sin \left( x \right)-\sin \left( x \right)\dfrac{d}{dx}\cos \left( x \right)}{{{\cos }^{2}}\left( x \right)}$
Now we know that $\dfrac{d}{dx}\sin \left( x \right)=\cos \left( x \right)$,and $\dfrac{d}{dx}\cos \left( x \right)=-\sin \left( x \right)$ therefore on substituting them in the expression, we get:
$\Rightarrow y'=\dfrac{\cos \left( x \right)\cos \left( x \right)-\sin \left( x \right)\left( -\sin \left( x \right) \right)}{{{\cos }^{2}}\left( x \right)}$
Now we know that the multiplication of two negative terms yields a positive term therefore, on simplifying the terms in the numerator, we get:
$\Rightarrow y'=\dfrac{{{\cos }^{2}}\left( x \right)+{{\sin }^{2}}\left( x \right)}{{{\cos }^{2}}\left( x \right)}$
Now we know the trigonometric identity that ${{\sin }^{2}}\left( x \right)+{{\cos }^{2}}\left( x \right)=1$ therefore on substituting, we get:
$\Rightarrow y'=\dfrac{1}{{{\cos }^{2}}\left( x \right)}$
Now we know that $\dfrac{1}{\cos \left( x \right)}=\sec \left( x \right)$ therefore, we get:
$\Rightarrow y'={{\sec }^{2}}x$, which is the required derivative.
Therefore, we can write:
$\Rightarrow \dfrac{d}{dx}\tan \left( x \right)={{\sec }^{2}}\left( x \right)$, which is the required solution.

Note: The solution found in this question should be remembered as a direct result or formula of the term $y=\tan \left( x \right)$ and substituted in questions wherever derivative of $\tan \left( x \right)$ is required. It is to be remembered that differentiation is the inverse of integration. If the derivative of a term $a$ is $b$, then the integration of the term $b$ will be $a$.