Question

How do you find the derivative of $\sin \left( \cos \left( 6x \right) \right)$?

Answer 1

This question is from the topic of calculus. In this question, we will find the derivative of $\sin \left( \cos \left( 6x \right) \right)$. In solving this question, we will first understand about the chain. After that, we will apply the chain rule to solve this question. After that, we will use some differentiation formulas to solve the further question. After solving the further question, we will get our answer.

Complete step by step solution:
Let us solve this question.
In this question, we were asked to find the derivative of $\sin \left( \cos \left( 6x \right) \right)$.
We will use chain rule here.
Let us first understand about the chain rule.
The chain rule in differentiation helps us differentiate the composite functions. The composite functions are in the form of $f\left( g\left( x \right) \right)$, where f and g are two different functions (or may be the same).
The chain rule says that if we have to do the differentiation of the composite function, then we will differentiate the composite function in the following way:
$\dfrac{d}{dx}\left[ f\left( g\left( x \right) \right) \right]=f'\left( g\left( x \right) \right)g'\left( x \right)$
So, we can differentiate the function $\sin \left( \cos \left( 6x \right) \right)$ using the chain rule. We can write
$\dfrac{d}{dx}\left[ \sin \left( \cos \left( 6x \right) \right) \right]$
We will use the formula: $\dfrac{d}{dx}\left( \sin x \right)=\cos x$ here. We can write
$\Rightarrow \dfrac{d}{dx}\left[ \sin \left( \cos \left( 6x \right) \right) \right]=\left[ \cos \left( \cos \left( 6x \right) \right) \right]\dfrac{d}{dx}\left( \cos \left( 6x \right) \right)$
Now, we will again use chain rule here. Taking ‘f’ as function of cos and ‘g’ as function 6x. So, using the formula: $\dfrac{d}{dx}\left( \cos x \right)=-\sin x$, we can write
$\Rightarrow \dfrac{d}{dx}\left[ \sin \left( \cos \left( 6x \right) \right) \right]=\left[ \cos \left( \cos \left( 6x \right) \right) \right]\left( -\sin x\left( 6x \right) \right)\dfrac{d}{dx}\left( 6x \right)$
$\Rightarrow \dfrac{d}{dx}\left[ \sin \left( \cos \left( 6x \right) \right) \right]=\left[ \cos \left( \cos \left( 6x \right) \right) \right]\left( -\sin \left( 6x \right) \right)\dfrac{d}{dx}\left( 6x \right)$
Using the formula: $\dfrac{d}{dx}\left( kx \right)=k$, where k is the constant. We can write
$\Rightarrow \dfrac{d}{dx}\left[ \sin \left( \cos \left( 6x \right) \right) \right]=\left[ \cos \left( \cos \left( 6x \right) \right) \right]\left( -\sin \left( 6x \right) \right)6$
We can write the above equation as
$\Rightarrow \dfrac{d}{dx}\left[ \sin \left( \cos \left( 6x \right) \right) \right]=-6\left[ \cos \left( \cos \left( 6x \right) \right) \right]\left( \sin \left( 6x \right) \right)$
Hence, we can say that the derivative of $\sin \left( \cos \left( 6x \right) \right)$ is $-6\left[ \cos \left( \cos \left( 6x \right) \right) \right]\left( \sin \left( 6x \right) \right)$

Note: We should have a better knowledge in the topic of calculus to solve this type of question easily.
We should know about the chain rule for differentiation. The chain rule is:
$\dfrac{d}{dx}\left[ f\left( g\left( x \right) \right) \right]=f'\left( g\left( x \right) \right)g'\left( x \right)$
Where, f and g are two functions.
We should remember the following formulas:
$\dfrac{d}{dx}\left( \sin x \right)=\cos x$
$\dfrac{d}{dx}\left( \cos x \right)=-\sin x$
$\dfrac{d}{dx}\left( kx \right)=k$
Where, k is any constant.