Question

How do you use the chain rule to differentiate $y={{\sin }^{3}}x+{{\cos }^{3}}x$?

Answer 1

In this question we have a composite function in the form $f\left( x \right)=g\left( h\left( x \right) \right)$ which has no direct formula for calculating the derivative therefore, we will use the chain rule on the function which is $f'\left( x \right)=g'\left( h\left( x \right) \right)h'\left( x \right)$. We have the expression in the form of addition of $2$ terms both of which are in the form of $g\left( h\left( x \right) \right)$. In both the terms we will consider $g\left( x \right)={{x}^{3}}$ and $h\left( x \right)=\sin x$ and $h\left( x \right)=\cos x$ respectively.

Complete step by step solution:
We have the expression given as:
$\Rightarrow y={{\sin }^{3}}x+{{\cos }^{3}}x$
Since we have to find the derivative of the term, it can be written as:
$\Rightarrow y'=\dfrac{d}{dx}\left( {{\sin }^{3}}x+{{\cos }^{3}}x \right)$
On splitting the derivative, we get:
$\Rightarrow y'=\dfrac{d}{dx}{{\sin }^{3}}x+\dfrac{d}{dx}{{\cos }^{3}}x\to \left( 1 \right)$
Now since there is no direct formula for calculating the derivative of the given expression, we will use the chain rule which is $f'\left( x \right)=g'\left( h\left( x \right) \right)h'\left( x \right)$.
Consider the first part of the equation $\left( 1 \right)$:
$\Rightarrow \dfrac{d}{dx}{{\sin }^{3}}x$
In this term, we have $g\left( x \right)={{x}^{3}}$ and $h\left( x \right)=\sin x$
We know that $g'\left( x \right)=\dfrac{d\left( {{x}^{3}} \right)}{dx}=3{{x}^{2}}$, and since $h\left( x \right)=\sin x$, we are using chain rule on both the terms, we will write it as:
$\Rightarrow 3{{\sin }^{2}}x\times h'\left( x \right)$
Now $h\left( x \right)=\sin x$
We know that $h'\left( x \right)=\dfrac{d}{dx}\left( \sin x \right)=\cos x$ therefore, on substituting, we get:
$\Rightarrow 3{{\sin }^{2}}x\cos x\to \left( 2 \right)$
Consider the second part of the equation $\left( 1 \right)$:
$\Rightarrow \dfrac{d}{dx}{{\cos }^{3}}x$
In this term, we have $g\left( x \right)={{x}^{3}}$ and $h\left( x \right)=\cos x$
We know that $g'\left( x \right)=\dfrac{d\left( {{x}^{3}} \right)}{dx}=3{{x}^{2}}$, and since $h\left( x \right)=\cos x$, we are using chain rule on both the terms, we will write it as:
$\Rightarrow 3{{\sin }^{2}}x\times h'\left( x \right)$
Now $h\left( x \right)=\sin x$
We know that $h'\left( x \right)=\dfrac{d}{dx}\left( \cos x \right)=-\sin x$ therefore, on substituting, we get:
$\Rightarrow 3{{\cos }^{2}}x\left( -\sin x \right)$
On simplifying, we get:
$\Rightarrow -3{{\cos }^{2}}x\sin x\to \left( 3 \right)$
On combining equations $\left( 2 \right)$ and $\left( 3 \right)$, we get:
$\Rightarrow y'=3{{\sin }^{2}}x\cos x-3{{\cos }^{2}}x\sin x$
Now we can see that the term $3\sin x\cos x$ is common in both the terms therefore, on taking it out as common, we get:
$\Rightarrow y'=3\sin x\cos x\left( \sin x-\cos x \right)$, which is the required solution.

Note: It is to be remembered that the chain rule is a different rule than the product and the quotient rule. Is to be remembered that there can be more than two functions in a composite function. The general rule to differentiate is by first solving the outer function and then moving to the inward functions.
It is to be also remembered that differentiation is the inverse of integration. If the integration of a term $a$ is $b$, then the differentiation of the term $b$ will be $a$ and vice versa.