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Question

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

How do you find the derivative of tan3(x){\tan ^3}\left( x \right)?

Explanation

Solution

In the given problem, we are required to differentiate tan3(x){\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 tan3(x){\tan ^3}\left( x \right) with respect to x will be done layer by layer. Also the derivative of tan(x)\tan (x) with respect to x must be remembered.

Complete step by step answer:
So, Derivative of tan3(x){\tan ^3}\left( x \right) with respect to x can be calculated as ddx(tan3(x))\dfrac{d}{{dx}}\left( {{{\tan }^3}\left( x \right)} \right) .
Now, ddx(tan3(x))\dfrac{d}{{dx}}\left( {{{\tan }^3}\left( x \right)} \right)
Now, Let us assume u=tanxu = \tan x. So substituting tanx\tan x as uu, we get,
== ddx(u3)\dfrac{d}{{dx}}\left( {{u^3}} \right)
Now, we know that the derivative of yn{y^n} with respect to y is nyn1n{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 tan3(x){\tan ^3}\left( x \right) with respect to xx is 3tan2xsec2x3{\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.