Question

Find the derivative of the following:
(i) $\sin x\cos x$

Answer 1

Hint: For solving this question we will use some standard results differentiation of $y=\sin x$ , $y=\cos x$ and then apply the product rule of differentiation to differentiate the given term with respect to $x$ correctly.

Complete step-by-step solution -
Given:
We have to find the derivative of $y=\sin x\cos x$ with respect to $x$ .
Now, before we proceed we should know the following formulas and concepts of trigonometry and differential calculus:
1. If $y=f\left( x \right)\cdot g\left( x \right)$ , then $\dfrac{dy}{dx}=\dfrac{d\left( f\left( x \right)\cdot g\left( x \right) \right)}{dx}={f}'\left( x \right)\cdot g\left( x \right)+f\left( x \right)\cdot {g}'\left( x \right)$. This is also known as the product rule of differentiation.
2. If $y=\sin x$ , then $\dfrac{dy}{dx}=\dfrac{d\left( \sin x \right)}{dx}=\cos x$.
3. If $y=\cos x$ , then $\dfrac{dy}{dx}=\dfrac{d\left( \cos x \right)}{dx}=-\sin x$.
Now, we will use the above-mentioned formulas and concepts to differentiate $y=\sin x\cos x$ with respect to $x$.
Now, let $y=\sin x\cos x=f\left( x \right)\cdot g\left( x \right)$ . Where, $f\left( x \right)=\sin x$ and $g\left( x \right)=\cos x$.
Now, as $f\left( x \right)=\sin x$ so, we can write its differentiation with respect to $x$ from the formula written in the second point. Then,
$\begin{align}
  & f\left( x \right)=\sin x \\
 & \Rightarrow {f}'\left( x \right)=\dfrac{d\left( \sin x \right)}{dx} \\
 & \Rightarrow {f}'\left( x \right)=\cos x....................\left( 1 \right) \\
\end{align}$
Now, as $g\left( x \right)=\cos x$ so, we can write its differentiation with respect to $x$ from the formula written in the above points. Then,
$\begin{align}
  & g\left( x \right)=\cos x \\
 & \Rightarrow {g}'\left( x \right)=\dfrac{d\left( \cos x \right)}{dx} \\
 & \Rightarrow {g}'\left( x \right)=-\sin x............\left( 2 \right) \\
\end{align}$
Now, as per our assumption $y=\sin x\cos x=f\left( x \right)\cdot g\left( x \right)$ so, we can use the product rule of differential calculus to write the derivative of $y=\sin x\cos x$ with respect to $x$ . Then,
$\begin{align}
  & y=\sin x\cos x=f\left( x \right)\cdot g\left( x \right) \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( \sin x\cos x \right)}{dx}=\dfrac{d\left( f\left( x \right)\cdot g\left( x \right) \right)}{dx} \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( \sin x\cos x \right)}{dx}={f}'\left( x \right)\cdot g\left( x \right)+f\left( x \right)\cdot {g}'\left( x \right) \\
\end{align}$
Now, substituting ${f}'\left( x \right)=\cos x$ from equation (1) and \[{g}'\left( x \right)=-\sin x\] from equation (2) in the above equation. Then,
$\begin{align}
  & \dfrac{dy}{dx}=\dfrac{d\left( \sin x\cos x \right)}{dx}={f}'\left( x \right)\cdot g\left( x \right)+f\left( x \right)\cdot {g}'\left( x \right) \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( \sin x\cos x \right)}{dx}=\cos x\cdot g\left( x \right)-\sin x\cdot f\left( x \right) \\
\end{align}$
Now, as per our assumption $f\left( x \right)=\sin x$ and $g\left( x \right)=\cos x$ . Then,
$\begin{align}
  & \dfrac{dy}{dx}=\dfrac{d\left( \sin x\cos x \right)}{dx}=\cos x\cdot g\left( x \right)-\sin x\cdot f\left( x \right) \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( \sin x\cos x \right)}{dx}=\cos x\cdot \cos x-\sin x\cdot \sin x \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( \sin x\cos x \right)}{dx}={{\cos }^{2}}x-{{\sin }^{2}}x \\
\end{align}$
Now, as we will use the formula $\cos 2\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta $ in the above equation. Then,
$\begin{align}
  & \dfrac{dy}{dx}=\dfrac{d\left( \sin x\cos x \right)}{dx}={{\cos }^{2}}x-{{\sin }^{2}}x \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( \sin x\cos x \right)}{dx}=\cos 2x \\
\end{align}$
Thus, $\dfrac{dy}{dx}=\cos 2x$ will be the derivative of a given function with respect to $x$

Note: Here, the student should know how to apply the product rule of differentiation to find the differentiation of functions of the form $y=f\left( x \right)\cdot g\left( x \right)$ . Moreover, we could also write $y=\sin x\cos x=\dfrac{\sin 2x}{2}$ and after that, we will find the derivative of the function $y=\dfrac{\sin 2x}{2}$ with respect to $x$ ,i.e. $\dfrac{dy}{dx}=\dfrac{d\left( \dfrac{\sin 2x}{2} \right)}{dx}=\dfrac{1}{2}\dfrac{d\left( \sin 2x \right)}{dx}$ . Then, we can write $\dfrac{1}{2}\dfrac{d\left( \sin 2x \right)}{dx}=\cos 2x$ to write the final answer correctly.