Question

How do you differentiate $f\left( x \right)={{\left( 7-8{{x}^{3}} \right)}^{2}}$ using the chain rule?

Answer 1

We need to have a sound knowledge in the differentiation topic in order to solve this question. We use what is known as the chain rule in differentiation in order to simplify the given function. This chain rule is represented as $\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)=f'\left( g\left( x \right) \right).g'\left( x \right).$ Using this, we can solve the given question.

Complete step by step solution:
We need to simplify the given function $f\left( x \right)={{\left( 7-8{{x}^{3}} \right)}^{2}}$ using the chain rule of differentiation.
We know the chain rule is defined as follows: The derivative of a composite function can be calculated by differentiating the outside function and the inner function and taking the product of the two. This can be given in the form of a formula as,
$\Rightarrow \dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)=f'\left( g\left( x \right) \right).g'\left( x \right)\ldots \ldots \left( 1 \right)$
Here, $f\left( g\left( x \right) \right)$ is the composite function, $f'\left( g\left( x \right) \right)$ is the differentiation or derivative of the outer function and $g'\left( x \right)$ is the derivative of the inner function.
Looking at the question, we can take $f\left( g\left( x \right) \right)={{\left( 7-8{{x}^{3}} \right)}^{2}}$ as the outer function and $g\left( x \right)=\left( 7-8{{x}^{3}} \right)$ as the inner function. Here, the outer and inner functions are related as follows,
$f\left( g\left( x \right) \right)=g{{\left( x \right)}^{2}}.$
Looking at the above formula represented by the equation (1), we need to find out the terms $f'\left( g\left( x \right) \right)$ and $g'\left( x \right).$ In order to do so, we need to differentiate the outer function first, followed by the inner function.
Differentiation of the inner function $g\left( x \right)=\left( 7-8{{x}^{3}} \right)$ is given by,
$\Rightarrow g'\left( x \right)=\left( 0-8\times 3{{x}^{3-1}} \right)$
This is done by using the basic differentiation formula $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}.$
Next, we simplify the equation by multiplying the terms and subtracting the power.
$\Rightarrow g'\left( x \right)=-24{{x}^{2}}$
The outer function can be represented in terms of the inner function as given above, that is,
$f\left( g\left( x \right) \right)=g{{\left( x \right)}^{2}}$
Therefore, substituting the value of $g\left( x \right),$
$\Rightarrow f'\left( g\left( x \right) \right)=2.g\left( x \right)$
Substituting these in equation (1),
$\Rightarrow \dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)=2.g\left( x \right).-24{{x}^{2}}$
We know that $g\left( x \right)=\left( 7-8{{x}^{3}} \right)$ and substituting this in the above equation,
$\Rightarrow \dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)=-2.24.{{x}^{2}}.\left( 7-8{{x}^{3}} \right)$
Multiplying the terms outside the bracket,
$\Rightarrow \dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)=-48{{x}^{2}}.\left( 7-8{{x}^{3}} \right)$
Now multiplying each term inside the bracket with the term outside,
$\Rightarrow \dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)=-336{{x}^{2}}+384{{x}^{5}}$
Hence, the final answer after differentiating $f\left( x \right)={{\left( 7-8{{x}^{3}} \right)}^{2}}$ using the chain rule is $-336{{x}^{2}}+384{{x}^{5}}.$

Note: To answer this question, students are required to know the chain rule in the topic of differentiation well. We can also solve this question without using the chain rule by expanding the equation and then differentiating normally. But since it is asked in question, it is best to proceed with the method shown in this solution.