Question

How do you find the derivative of ${\tan ^3}\left( x \right)$?

Answer 1

Hint: In the given problem, we are required to differentiate ${\tan ^3}\left( x \right)$ with respect to x. The given function is a composite function, so we will have to apply the chain rule of differentiation in the process of differentiation. So, differentiation of ${\tan ^3}\left( x \right)$ with respect to x will be done layer by layer. Also the derivative of $\tan (x)$ with respect to x must be remembered.

Complete step by step answer:
So, Derivative of ${\tan ^3}\left( x \right)$ with respect to x can be calculated as $\dfrac{d}{{dx}}\left( {{{\tan }^3}\left( x \right)} \right)$ .
Now, $\dfrac{d}{{dx}}\left( {{{\tan }^3}\left( x \right)} \right)$
Now, Let us assume $u = \tan x$. So substituting $\tan x$ as $u$, we get,
$ = $$\dfrac{d}{{dx}}\left( {{u^3}} \right)$
Now, we know that the derivative of ${y^n}$ with respect to y is \[n{y^{n - 1}}\].
So, we get,
$ = $\[3{u^{3 - 1}}\left( {\dfrac{{du}}{{dx}}} \right)\]
Simplifying further, we get,
$ = $\[3{u^2}\left( {\dfrac{{du}}{{dx}}} \right)\]
Now, substituting the value of u in terms of x as $\tan x$, we get,
$ = $\[3{\left( {\tan x} \right)^2} \times \dfrac{{d\left( {\tan x} \right)}}{{dx}}\]
Derivative of tangent function $\tan y$ with respect to y is ${\sec ^2}y$. So, we have,
$ = $\[3{\tan ^2}x{\sec ^2}x\]
So, the derivative of ${\tan ^3}\left( x \right)$ with respect to $x$ is \[3{\tan ^2}x{\sec ^2}x\].

Note:
The derivatives of basic trigonometric functions must be learned by heart in order to find derivatives of complex composite functions using chain rule of differentiation. The chain rule of differentiation involves differentiating a composite by introducing new unknowns to ease the process and examine the behaviour of function layer by layer.