Question

Find the derivative of the following function:
$\sin x\cos x$

Answer 1

Hint: In this question apply the product rule of differentiation which is given as $\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{d}{{dx}}v + v\dfrac{d}{{dx}}u$ later on in the solution apply the differentiation property of sin x and cos x which is given as $\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x{\text{ and }}\dfrac{d}{{dx}}\left( {\cos x} \right) = -\sin x$ so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Let
$y = \sin x\cos x$
Now differentiate it w.r.t. x we have,
$ \Rightarrow \dfrac{d}{{dx}}y = \dfrac{d}{{dx}}\left[ {\sin x\cos x} \right]$
Now here we use product rule of differentiate which is given as
$\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{d}{{dx}}v + v\dfrac{d}{{dx}}u$ so use this property in above equation we have,
$ \Rightarrow \dfrac{d}{{dx}}y = \sin x\dfrac{d}{{dx}}\left( {\cos x} \right) + \left( {\cos x} \right)\dfrac{d}{{dx}}\left( {\sin x} \right)$
Now as we know that differentiation of $\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x{\text{ and }}\dfrac{d}{{dx}}\left( {\cos x} \right) = -\sin x$ so use this property in above equation we have,
$ \Rightarrow \dfrac{d}{{dx}}y = \sin x\left( { - \sin x} \right) + \left( {\cos x} \right)\left( {\cos x} \right)$
Now simplify this equation we have,
$ \Rightarrow \dfrac{d}{{dx}}y = {\cos ^2}x - {\sin ^2}x$
Now as we know that ${\cos ^2}x - {\sin ^2}x = \cos 2x$
$ \Rightarrow \dfrac{d}{{dx}}y = \cos 2x$
So this is the required differentiation.

Note – Whenever we face such types of questions the key concept is always recall the formula of product rule of differentiation, formula of sin x and cos x differentiation which is stated above then first apply the product rule as above then use the property of differentiation of sin x and cos x as above and simplify we will get the required answer.