Question

Find the derivative of $\sin2x \cos3x$.

Answer 1

To solve this question, firstly we will proceed using the product rule of differentiation of two functions.
Later on, we will use the Chain Rule of differentiation to differentiate the smaller parts. After doing the above steps, then we will do some rearrangements of terms and hence, the differentiation of $\sin2x \cos3x$ will be equal to the final answer.

Complete step-by-step solution:
We are to find the derivative of $\sin2x \cos3x$, i.e. $\dfrac{d}{dx}\left( \sin 2x\cos 3x \right)$
At first, we differentiate using the Product Rule which states
$\dfrac{d}{dx}\left[ f\left( x \right)g\left( x \right) \right]=\left[ f\left( x \right)\times \dfrac{d}{dx}g\left( x \right) \right]+\left[ g\left( x \right)\times \dfrac{d}{dx}f\left( x \right) \right]$
Here $f(x)= \sin(2x)$ and $g(x)= \cos(3x)$ and we know that $\dfrac{d}{dx}(\sin ax)=a\cos ax$ and$\dfrac{d}{dx}(\cos ax)=-a\sin ax$.
∴$\dfrac{d}{dx}\left( \sin 2x\cos 3x \right)$=$\sin(2x)\dfrac{d}{dx}[\cos(3x)]+\cos(3x)\dfrac{d}{dx}[\sin(2x)]$
Now, we will use Chain rule to differentiate further.
The chain rule is given by $\dfrac{d}{dx}f\left( g(x) \right)=f'\left( g(x) \right)\times g'(x)$
Therefore, applying this rule, we get

& \dfrac{d}{dx}[\cos(3x)] \\\ & =-\sin 3x.\dfrac{d}{dx}(3x) \\\ & =-3\sin 3x \\\ \end{aligned}$$ And, $$\begin{aligned} & \dfrac{d}{dx}[\sin(2x)] \\\ & =\cos 2x.\dfrac{d}{dx}(2x) \\\ & =2\cos 2x \\\ \end{aligned}$$ Therefore, $\dfrac{d}{dx}\left( \sin 2x\cos 3x \right)$ $$\begin{aligned} & = \sin(2x)\dfrac{d}{dx}[\cos(3x)]+cos(3x)\dfrac{d}{dx}[\sin(2x)] \\\ & =\sin 2x.(-3\sin 3x)+\cos 3x(2\cos 2x) \\\ & =2\cos 2x\cos 3x-3\sin 2x\sin 3x \\\ \end{aligned}$$ **Hence, the derivative of $\sin2x \cos3x$ is $$2\cos 2x\cos 3x-3\sin 2x\sin 3x$$** **Note:** Note that, the Product Rule of differentiation is $$\dfrac{d}{dx}\left[ f\left( x \right)g\left( x \right) \right]=\left[ f\left( x \right)\times \dfrac{d}{dx}g\left( x \right) \right]+\left[ g\left( x \right)\times \dfrac{d}{dx}f\left( x \right) \right]$$, and the Chain Rule for differentiation is $\dfrac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)$. Also, remember that $\dfrac{d}{dx}(\sin ax)=a\cos ax$ and $\dfrac{d}{dx}(cosax)=-a\sin ax$. Try not to make any calculation mistakes.