What is the derivative of $\sin (2x)\cos(3x)?$
Is it
$\cos (2x) \cos(3x)-\sin (2x) \sin(3x)$
or
$2\cos (2x) \cos(3x)-3\sin (2x) \sin(3x)?$
The answer is given the latter but I only get the former.
Please offer your assistance for the same.
Thankyou
4 Answers
$\begingroup$You've noticed that you need to use the product rule, but don't forget to use chain rule! For example,
$$\bigl(\sin(f(x))\bigr)'=\cos(f(x))\cdot f'(x)$$
$\endgroup$ $\begingroup$Hint
Complete $$(\cos(ax))'=\cdots\quad;\quad (\sin(ax))'=\cdots$$ and : $$(f\times g)'=\cdots$$ Apply these results.
Answer
$\endgroup$ 8 $\begingroup$$(\cos(ax))'=-a\sin(ax)\quad;\quad (\sin(ax))'=a\cos(ax)\quad;\quad (f\times g)'=f'g+g'f$ so we apply: $(\sin(2x)\cos(3x))'=(\sin(2x))'\cos(3x)+\sin(2x)(\cos(3x))'=2\cos (2x) \cos(3x)-3\sin (2x) \sin(3x)$
The product rule states: $$\frac d {dx}(f(x)\times g(x))=f'(x)\times g(x) + g'(x)\times f(x)$$ So let f(x) = sin(2x) and g(x) = cos (3x). You should be familiar with the derivative of sin(ax) and cos(ax), so just plug those in.
Useful:
$$\frac d {dx} \sin(ax) = a\cos(ax)$$ $$\frac d {dx} \cos(ax) = -a\sin(ax)$$
$\endgroup$ 1 $\begingroup$HINT:
Why don't we use $$2\sin B\cos A=\sin(A+B)-\sin(A-B)$$ to find $$2\sin2A\cos3A=\sin5A-\sin A$$
$\endgroup$
