Question

What is the derivative of $y=3\sin x-\sin 3x$ ?

Answer 1

Here in this question we have been asked to find the derivative of $y=3\sin x-\sin 3x$ for answering this question we will use the formula given as $\dfrac{d}{dx}\sin x=\cos x$ and $\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}$ .

Complete step-by-step answer:
Now considering from the question we have been asked to find the derivative of $y=3\sin x-\sin 3x$.
We can simply write the given expression as
$\begin{aligned} & \dfrac{dy}{dx}=\dfrac{d}{dx}\left( 3\sin x-\sin 3x \right) \\\ & \Rightarrow 3\dfrac{d}{dx}\cos x-\dfrac{d}{dx}\sin 3x \\\ \end{aligned}$ .
From the basic concepts of derivations we know that the formula for finding the derivative of $\sin x$ is given as $\cos x$ .
Similarly we are also aware of the chain rule given as $\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}$ .
Now by using these concepts we can simply write this expression as $\Rightarrow 3\cos x-\dfrac{d}{dx}\sin 3x$ .
Now we will assume $u=3x$ then we will have $\Rightarrow 3\cos x-\dfrac{d}{dx}\sin u$ .
Now by applying the chain rule we will have $\Rightarrow 3\cos x-\dfrac{d}{du}\sin u\left( \dfrac{du}{dx} \right)$ .
Now by applying the formula of sine derivative we will have $\Rightarrow 3\cos x-\cos u\left( \dfrac{du}{dx} \right)$ .
Now by substituting $u=3x$ we will have $\Rightarrow 3\cos x-\cos 3x\left( \dfrac{d}{dx}3x \right)$ .
Now by further simplifying this we will have $\Rightarrow 3\cos x-3\cos 3x$ .
Therefore we can conclude that the derivative of $y=3\sin x-\sin 3x$ will be given as $3\cos x-3\cos 3x$ .

Note: While answering questions of this type we should be very sure with the concepts that we are going to use in between the steps. Someone can make a mistake unintentionally if they consider the simplification in between the steps in a wrong way like for an example if we consider $\Rightarrow 3\cos x-\cos 3x\left( \dfrac{d}{dx}3x \right)=\Rightarrow 3\cos x-\cos 3x\left( 4 \right)$ which will lead us to end up having a wrong conclusion.