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 ${\displaystyle \frac{dy}{dx}}$. 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${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{df(g(x))}{dx}}={\displaystyle \frac{df(g(x))}{dg(x)}}\times {\displaystyle \frac{dg(x)}{dx}}$.
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
$\Rightarrow {\displaystyle \frac{dy}{dx}}={\displaystyle \frac{df(g(x))}{dx}}={\displaystyle \frac{df(g(x))}{dg(x)}}\times {\displaystyle \frac{dg(x)}{dx}}$
$\Rightarrow {\displaystyle \frac{d\sin (4x)}{dx}}={\displaystyle \frac{d\sin (4x)}{d(4x)}}\times {\displaystyle \frac{d(4x)}{dx}}$
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
$\Rightarrow {\displaystyle \frac{d\sin (4x)}{dx}}=\cos (4x)\times {\displaystyle \frac{d(4x)}{dx}}$
Now for differentiating $4x$, we will take the constant out, i.e. $4$ and use one of the rule of differentiation that states ${\displaystyle \frac{d{x}^n}{dx}}=n{x}^{n-1}$. Here $n=1$, so ${\displaystyle \frac{d{x}^1}{dx}}=1\times x^{1-1}=x^0=1$. On substituting these values, we will get
$\Rightarrow {\displaystyle \frac{d\sin (4x)}{dx}}=\cos (4x)\times 4\times {\displaystyle \frac{d(x)}{dx}}$
$\Rightarrow {\displaystyle \frac{d\sin (4x)}{dx}}=\cos (4x)\times 4\times 1$
On further simplifying, we will get
$\Rightarrow {\displaystyle \frac{d\sin (4x)}{dx}}=4\cos (4x)$

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 ${\displaystyle \frac{dy}{dx}}$. But it can also be expressed as $Dy$ as $D$ is sometimes used in place of ${\displaystyle \frac{d}{dx}}$.