How do you differentiate $f\left( x \right) = \tan \left( {5x} \right)$?

Hint: We can use chain rule to differentiate $f\left( x \right) = \tan \left( {5x} \right)$. For this, first find the differentiation of $5x$ with respect to $x$. Then, find the differentiation of $\tan \left( {5x} \right)$ with respect to $5x$. Multiply these and use chain rule to get the required derivative.

Formula used:
Chain Rule:
Chain rule is applied when the given function is the function of function i.e.,
if y is a function of x, then $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dx}}$ or $\dfrac{{dy}}{{dx}} = \dfrac{{dy}}{{du}} \times \dfrac{{du}}{{dv}} \times \dfrac{{dv}}{{dx}}$.
The differentiation of the product of a constant and a function = the constant $ \times $ differentiation of the function.
i.e., $\dfrac{d}{{dx}}\left( {kf\left( x \right)} \right) = k\dfrac{d}{{dx}}\left( {f\left( x \right)} \right)$, where $k$ is a constant.
The differentiation of tangent function is square of secant function.
i.e., $\dfrac{d}{{dx}}\left( {\tan x} \right) = {\sec ^2}x$

Complete step by step answer:
We have to find the derivative of $f\left( x \right) = \tan \left( {5x} \right)$.
Here, $f\left( x \right) = \tan g\left( x \right)$, where $g\left( x \right) = 5x$.
We have to find the differentiation of $f$ with respect to $x$.
It can be done using Chain Rule.
\[\dfrac{{df}}{{dx}} = \dfrac{{df}}{{dg}} \times \dfrac{{dg}}{{dx}}\]…(1)
i.e., Differentiation of $f$ with respect to $x$ is equal to product of differentiation of $f$ with respect to $g$, and differentiation of $g$ with respect to $x$.
We will first find the differentiation of $g$ with respect to $x$.
Here, $g\left( x \right) = 5x$
Differentiating $g$ with respect to $x$.
$\dfrac{{dg}}{{dx}} = \dfrac{d}{{dx}}\left( {5x} \right)$
Now, using the property that the differentiation of the product of a constant and a function = the constant $ \times $ differentiation of the function.
i.e., $\dfrac{d}{{dx}}\left( {kf\left( x \right)} \right) = k\dfrac{d}{{dx}}\left( {f\left( x \right)} \right)$, where $k$ is a constant.
So, in above differentiation, constant $5$ can be taken outside the differentiation.
$ \Rightarrow \dfrac{{dg}}{{dx}} = 5\dfrac{d}{{dx}}\left( x \right)$
Now, using the differentiation formula $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}},n \ne - 1$ in above differentiation, we get
$ \Rightarrow \dfrac{{dg}}{{dx}} = 5$…(2)
Now, we will find the differentiation of$f$ with respect to $g$.
Here, $f\left( x \right) = \tan g\left( x \right)$
Differentiating$f$ with respect to $g$.
$\dfrac{{df}}{{dg}} = \dfrac{d}{{dx}}\left( {\tan g\left( x \right)} \right)$
Now, using the differentiation formula $\dfrac{d}{{dx}}\left( {\tan x} \right) = {\sec ^2}x$ in above differentiation, we get
$ \Rightarrow \dfrac{{df}}{{dg}} = {\sec ^2}g\left( x \right)$…(2)
Put the value of $g\left( x \right)$ in the above equation.
Since, $g\left( x \right) = 5x$.
So, $\dfrac{{df}}{{dg}} = {\sec ^2}\left( {5x} \right)$…(3)
Put the value of $\dfrac{{df}}{{dg}},\dfrac{{dg}}{{dx}}$ from equation (2) and (3) in equation (1).
$\dfrac{{df}}{{dx}} = {\sec ^2}\left( {5x} \right) \times 5$
Multiplying the terms, we get
$ \Rightarrow \dfrac{{df}}{{dx}} = 5{\sec ^2}\left( {5x} \right)$

Therefore, the derivative of $f\left( x \right) = \tan \left( {5x} \right)$ is $f'\left( x \right) = 5{\sec ^2}\left( {5x} \right)$.

Note: Chain rule, in calculus, basic method for differentiating a composite function. If $f\left( x \right)$ and $g\left( x \right)$ are two functions, the function $f\left( {g\left( x \right)} \right)$ is calculated for a value of $x$ by first evaluating $g\left( x \right)$ and then evaluating the function $f$ at this value of $g\left( x \right)$, thus “chaining” the results together.