Question

How do you find the derivative of $\left( {{\sin }^{2}}x \right)\left( {{\cos }^{3}}x \right)$?

Answer 1

Hint: As the given function is the product of two functions so first we will use the product formula which is given by $\dfrac{d}{dx}\left( uv \right)=u\dfrac{d}{dx}v+v\dfrac{d}{dx}u$. Also the given function is a composite function so we will use the chain rule to find the derivative. We will also use power rule $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n\times {{x}^{n-1}}$, sine formula $\dfrac{d}{dx}(\sin x)=\cos x$and cos formula $\dfrac{d}{dx}(\cos x)=-\sin x$ to solve further.

Complete step-by-step solution:
We have been given a function $\left( {{\sin }^{2}}x \right)\left( {{\cos }^{3}}x \right)$.
We have to find a derivative of the given function.
Let us assume that $y=\left( {{\sin }^{2}}x \right)\left( {{\cos }^{3}}x \right)$
Now, the given function is the product of two functions so first we apply the product formula which is given as $\dfrac{d}{dx}\left( uv \right)=u\dfrac{d}{dx}v+v\dfrac{d}{dx}u$.
Differentiating the given function with respect to x and applying the formula we will get
$\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{\sin }^{2}}x \right)\left( {{\cos }^{3}}x \right) \\
 & \Rightarrow \dfrac{dy}{dx}=\left( {{\sin }^{2}}x \right)\dfrac{d}{dx}\left( {{\cos }^{3}}x \right)+\left( {{\cos }^{3}}x \right)\dfrac{d}{dx}\left( {{\sin }^{2}}x \right) \\
\end{align}$
Now, the functions $\left( {{\sin }^{2}}x \right)$ and $\left( {{\cos }^{3}}x \right)$are composite functions so we apply the chain rule.
Also we know that $\dfrac{d}{dx}\left( {{x}^{n}} \right)=n\times {{x}^{n-1}}$, $\dfrac{d}{dx}(\sin x)=\cos x$ and $\dfrac{d}{dx}(\cos x)=-\sin x$
So applying the above formulas and substituting the values we will get
$\Rightarrow \dfrac{dy}{dx}=\left( {{\sin }^{2}}x \right)3{{\cos }^{2}}x\left( -\sin x \right)+\left( {{\cos }^{3}}x \right)2\sin x\left( \cos x \right)$
Now, simplifying the above obtained equations we will get
$\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=-3{{\sin }^{3}}x{{\cos }^{2}}x+2{{\cos }^{4}}x\sin x \\
 & \Rightarrow \dfrac{dy}{dx}=2{{\cos }^{4}}x\sin x-3{{\sin }^{3}}x{{\cos }^{2}}x \\
\end{align}$
Hence we get the derivative of $\left( {{\sin }^{2}}x \right)\left( {{\cos }^{3}}x \right)$ as $2{{\cos }^{4}}x\sin x-3{{\sin }^{3}}x{{\cos }^{2}}x$.

Note: Here in this question we use both product rule and chain rule. Both are different so don’t be confused and it is necessary to have a clear idea about both. We cannot directly differentiate the product of functions or a composite function. The question is lengthy so be careful while applying the formula and be careful while doing calculations.