Question

How do use the chain rule to differentiate $y = {\cos ^6}x?$

Answer 1

Hint: As we know that the chain rule is used to differentiate the composite functions. The chain rule states that the derivative of $f\left( {g(x)} \right)$ is $f'\left( {g(x)} \right) \cdot g'(x)$. In this question, we have to differentiate the given equation of the trigonometric ratio using the chain rule. Here we will take the derivative of the inner function and then solve it.

Complete step by step answer:
As per the given question we have $y = {\cos ^6}x$.
Here we will first take the derivative as the normal i.e. $6 \cdot \cos^5 {(x)}$. The constant term here is $6$.
Now we take the derivative of the inner function i.e. cosine, so the derivative of $\cos (x)$ is$ - \sin (x)$.
By multiplying all the terms together we have: $6 \cdot \cos^5 {(x)} \times - \sin (x)$. It can be written as $ - 6\sin (x)\cos^5 {(x)}$.

Hence the required answer is $ - 6\sin (x)\cos^5 {(x)}$.

Note: Before solving this kind of question, we should be aware of all the basic differentiation rules in chain rule of differentiation. The difference rule says the derivative of a difference of functions is the difference of their derivatives. We should remember the basic derivatives such as $\dfrac{d}{{dx}}(\cos x) = - \sin x$, $\dfrac{d}{{dx}}(\sin x) = \cos x$, $\dfrac{d}{{dx}}(\tan x) = {\sec ^2}x$, $\dfrac{d}{{dx}}(\sec x) = \sec x\tan x$ and so on.