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Question

Question: What is the derivative of \({\tan ^4}\left( {3x} \right)\)?...

What is the derivative of tan4(3x){\tan ^4}\left( {3x} \right)?

Explanation

Solution

In the given problem, we are required to differentiate tan4(3x){\tan ^4}\left( {3x} \right) with respect to xx. 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 tan4(3x){\tan ^4}\left( {3x} \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 tan4(3x){\tan ^4}\left( {3x} \right) with respect to xx can be calculated as ddx(tan4(3x))\dfrac{d}{{dx}}\left( {{{\tan }^4}\left( {3x} \right)} \right). Now, ddx(tan4(3x))\dfrac{d}{{dx}}\left( {{{\tan }^4}\left( {3x} \right)} \right). Now, Let us assume u=tan(3x)u = \tan \left( {3x} \right). So substituting tan(3x)\tan \left( {3x} \right) as uu, we get,
ddx(u4)\Rightarrow \dfrac{d}{{dx}}\left( {{u^4}} \right)
Now, we know that the derivative of xn{x^n} with respect to xx is nxn1n{x^{n - 1}}. So, we get,
4u3×dudx\Rightarrow 4{u^3} \times \dfrac{{du}}{{dx}}
Now, putting back uuas tan(3x)\tan \left( {3x} \right), we get,
4(tan(3x))3×d(tan(3x))dx\Rightarrow 4{\left( {\tan \left( {3x} \right)} \right)^3} \times \dfrac{{d\left( {\tan \left( {3x} \right)} \right)}}{{dx}}
Now, let t=(3x)t = \left( {3x} \right). Then, doing the substitution, we get,
4(tan(3x))3×d(tant)dx\Rightarrow 4{\left( {\tan \left( {3x} \right)} \right)^3} \times \dfrac{{d\left( {\tan t} \right)}}{{dx}}
Derivative of tangent function tanx\tan x with respect to x is sec2x{\sec ^2}x. So, we have,
4(tan(3x))3×sec2t×dtdx\Rightarrow 4{\left( {\tan \left( {3x} \right)} \right)^3} \times {\sec ^2}t \times \dfrac{{dt}}{{dx}}
Now, putting back t as (3x)\left( {3x} \right), we get,
4(tan(3x))3×sec2(3x)×d(3x)dx\Rightarrow 4{\left( {\tan \left( {3x} \right)} \right)^3} \times {\sec ^2}\left( {3x} \right) \times \dfrac{{d\left( {3x} \right)}}{{dx}} because dtdx=d(3x)dx\dfrac{{dt}}{{dx}} = \dfrac{{d\left( {3x} \right)}}{{dx}}
We use the power rule of differentiation d(xn)dx=nxn1\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}} again. So, we get,
4(tan(3x))3×sec2(3x)×3\Rightarrow 4{\left( {\tan \left( {3x} \right)} \right)^3} \times {\sec ^2}\left( {3x} \right) \times 3
Simplifying the product of terms, we get,
12tan3(3x)sec2(3x)\therefore 12{\tan ^3}\left( {3x} \right){\sec ^2}\left( {3x} \right)

Hence, the derivative of tan4(3x){\tan ^4}\left( {3x} \right) with respect to xx is 12tan3(3x)sec2(3x)12{\tan ^3}\left( {3x} \right){\sec ^2}\left( {3x} \right).

Note: This type of composite function will be easily solved by the 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. We should always substitute back the value of variables in order to reach the final required answer.