Question

How do you find the derivative of $y = 2\cos x\sin x$?

Answer 1

Hint: This can be solved in two ways. Either we can use the normal product rule for differentiation or we can use the trigonometric identity $2\cos x\sin x = \sin 2x$.

Formula used:
Product rule of differentiation,
 $\dfrac{{d(uv)}}{{dx}} = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}$ , where $u,v$ are functions of $x$ .
Trigonometric identity,
${\cos ^2}x - {\sin ^2}x = \cos 2x$
Chain rule of differentiation,
$h {’}(x) = f {’}(g(x)) \times g {’}(x)$ , where $f,g$ are functions of $x$ and $h$ is composite function of $f,g$.

Complete step-by-step answer:
Let us try the product rule of differentiation to solve the problem.
Let $y = f(x) = 2\cos x\sin x$ and $g(x) = \cos x\sin x$
 $ \Rightarrow f ' (x) = 2g ' (x)$ ------------(1)
Therefore, let us find the derivative of $g(x)$ using the product rule of differentiation.
Here, let $u = \cos x$ and $v = \sin x$.
$ \Rightarrow \dfrac{{dv}}{{dx}} = \dfrac{{d(\sin x)}}{{dx}} = \cos x$ and
 $\dfrac{{du}}{{dx}} = \dfrac{{d(\cos x)}}{{dx}} = - \sin x$
Substituting, we get
$\dfrac{{d(uv)}}{{dx}} = \dfrac{{d(\cos x \times \sin x)}}{{dx}} = g ' (x)$
$v\dfrac{{du}}{{dx}} + u\dfrac{{dv}}{{dx}} = \cos x \times \cos x + \sin x \times - \sin x$
Equating,
$g ' (x) = {\cos ^2}x - {\sin ^2}x$
We know,
${\cos ^2}x - {\sin ^2}x = \cos 2x$
$ \Rightarrow g ' (x) = \cos 2x$
Therefore, from equation (1) we have,
$f ' (x) = 2\cos 2x$
Thus, the derivative of $y = 2\cos x\sin x$ is $2\cos 2x$.

Additional information:
 Let $f(x)$and $g(x)$ be two functions of $x$. Let $h(x)$ be another function such that $h(x) = f(g(x))$.
Also let $f ' (x),g ' (x),h ' (x)$ be the derivatives of those functions. Then, as per chain rule, we have
$h ' (x) = f ' (g(x)) \times g ' (x)$.

Note: This question can also be solved using the trigonometric identity $2\cos x\sin x = \sin 2x$
$ \Rightarrow y = f(x) = 2\cos x\sin x = \sin 2x$
Now, we find the derivative of $f(x)$ using standard derivatives as well as chain rule of differentiation.
The derivative of $\sin x$ is $\cos x$ , as we all know.
Also, the derivative of $2x$ is $2$.
Therefore, the derivative of $y = f(x) = 2\cos x\sin x = \sin 2x$ is
$\dfrac{{dy}}{{dx}} = \dfrac{{d(\sin 2x)}}{{dx}} = \cos 2x \times \dfrac{{d(2x)}}{{dx}}$
$ = \cos 2x \times 2 = 2\cos 2x$
Which is the same result we got using the previous method.