Question

What is the derivative of \[\sin 5x\]?

Answer 1

Hint: We have to find the derivative of sin 5x. This looks like a composition of two functions. One is a trigonometric function and the other is an algebraic function. So, we have to use the chain rule here. First, we will find the derivative of sin 5x, which is the outer function and then we will find the derivative of 5x which is the inner function. Once we have them, we will write the product of them to get the answer.

Complete step by step solution:
Let us first learn the concepts required. In order to solve this, let us follow the chain rule.
Chain rule is used to identify the inside and outside function. The general chain rule is that \[h(x)=f \left( g (x) \right)\]. The functions with logarithmic or exponential are considered to be as outside function and the rest of the function is considered as inside function.
In our case, $sin 5x$ is outside function and $5x$ is inside function. The chain rule states the sine function as \[\dfrac{d}{dx}[\sin u]=\cos u\cdot \dfrac{du}{dx}\].
Now let us solve the problem.
First, we will find the derivative of sin x. As we know that the derivative of \[\sin x\] is \[\cos x\].
\[\dfrac{d}{dx}\left( \sin x \right)=\cos x\]
Now substitute this derivative into the function we have. Then we get,
\[\dfrac{d}{dx}\left( \sin 5x \right)=\cos 5x\]
The next step is to multiply the function by the derivative of the inside function.
The derivative of the inside function is \[\dfrac{d}{dx}\left( 5x \right)=5\].
After multiplying we get,
Derivative of \[5x\] is
\[\dfrac{d}{dx}\left( 5x \right)=5\]
Hence the derivative of the given function is-
\[\dfrac{d}{dx}\left( \sin 5x \right)=5\cos 5x\]

Note: Whenever the function with respect to be derived is not given, then derivate it with respect to \[x\] for our convenience. Do not try to solve it by applying logarithmic functions because it can make this simple problem complicated. It can also be solved by chain rule and product rule.