Question

How do you differentiate $y = \sin (4x)$?

Answer 1

Hint: To differentiate means to find the derivative of a function or rate of change of a function with respect to some variable. If a function $y$ is to be differentiated with respect to $x$, then it will be written as $\dfrac{{dy}}{{dx}}$. In the question $\sin (4x)$ is a composite function. If $f(x)$ and $g(x)$ are two functions, then $f(g(x))$ is said to be a composite function. For the given question, $f(x) = \sin x$ and $g(x) = 4x$ such that $f(g(x)) = \sin (4x)$. To differentiate $\sin (4x)$, we will use the chain rule which is used to differentiate composite functions. The chain rule states that if $y = f(g(x))$, then$\dfrac{{dy}}{{dx}} = \dfrac{{df(g(x))}}{{dx}} = \dfrac{{df(g(x))}}{{dg(x)}} \times \dfrac{{dg(x)}}{{dx}}$.
Also, it must be known that the differentiation or derivative of $\sin x$ is $\cos x$.

Complete step by step solution:
It is given that $y = \sin (4x)$.
Here $f(g(x)) = \sin (4x)$ where $f(x) = \sin x$ and $g(x) = 4x$. On substituting these values in the chain rule of differentiation, we will get
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{df(g(x))}}{{dx}} = \dfrac{{df(g(x))}}{{dg(x)}} \times \dfrac{{dg(x)}}{{dx}}$
$ \Rightarrow \dfrac{{d\sin (4x)}}{{dx}} = \dfrac{{d\sin (4x)}}{{d(4x)}} \times \dfrac{{d(4x)}}{{dx}}$
It must be known that the differentiation or derivative of $\sin x$ with respect to $x$ is $\cos x$ where $x$ is the argument of the trigonometric function. Therefore the derivative of $\sin (4x)$ with respect to $4x$ is $\cos (4x)$. On substituting it, we will get
$ \Rightarrow \dfrac{{d\sin (4x)}}{{dx}} = \cos (4x) \times \dfrac{{d(4x)}}{{dx}}$
Now for differentiating $4x$, we will take the constant out, i.e. $4$ and use one of the rule of differentiation that states $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$. Here $n = 1$, so $\dfrac{{d{x^1}}}{{dx}} = 1 \times {x^{1 - 1}} = {x^0} = 1$. On substituting these values, we will get
$ \Rightarrow \dfrac{{d\sin (4x)}}{{dx}} = \cos (4x) \times 4 \times \dfrac{{d(x)}}{{dx}}$
$ \Rightarrow \dfrac{{d\sin (4x)}}{{dx}} = \cos (4x) \times 4 \times 1$
On further simplifying, we will get
$ \Rightarrow \dfrac{{d\sin (4x)}}{{dx}} = 4\cos (4x)$

Hence, when we differentiate $y = \sin (4x)$ we get $4\cos (4x)$ as the answer.

Note:
If a function $y$ is to be differentiated with respect to $x$, then it will be written as $\dfrac{{dy}}{{dx}}$. But it can also be expressed as $Dy$ as $D$ is sometimes used in place of $\dfrac{d}{{dx}}$.