Question

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

Answer 1

Hint: We are given a function, ${{\cos }^{2}}\left( 3x \right)$ and we have to find its derivative. We can see that our function is made by composition of different functions. So, we will use the chain rule to find its derivative. We will learn about chain rule and use it to find the derivative of ${{\cos }^{2}}\left( 3x \right)$. We will use $\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$, derivative of $\left( kx \right)$ is $k$ times derivative of $x$ to get our required answer.

Complete step by step solution:
We are given a function ${{\cos }^{2}}\left( 3x \right)$. We can see that it is not a simple function, it consists of more than one function. Our given function is made by the composition of $\cos x,3x$ and ${{x}^{2}}$. We know that when a function is made by the composition of different functions, then in order to find its derivative, we use the chain rule. The chain rule states that if we have a function as $u\left[ v\left( x \right) \right]$, then,
$\dfrac{du\left[ v\left( x \right) \right]}{dx}={{u}^{'}}\left[ v\left( x \right) \right].{{v}^{'}}\left( x \right)$
Now in our function, we have, ${{\cos }^{2}}\left( 3x \right)$, so it is written as,
${{\cos }^{2}}\left( 3x \right)={{\left[ \cos \left( 3x \right) \right]}^{2}}$
Now, we know that,
$\dfrac{d\left( {{x}^{n}} \right)}{dx}=n{{x}^{n-1}}$
So, putting $\left[ \cos \left( 3x \right) \right]$ as $y$, we get,
$\begin{align}
  & \dfrac{d\left( {{y}^{2}} \right)}{dx}=2{{y}^{2-1}} \\
 & \Rightarrow \dfrac{d\left( {{y}^{2}} \right)}{dx}=2\left[ \cos \left( 3x \right) \right] \\
\end{align}$
We know that the derivative of $\cos x=-\sin x$, so the derivative of $\cos \left( 3x \right)=-\sin \left( 3x \right)$.
And lastly the derivative of $3x$ is,
$\dfrac{d\left( 3x \right)}{dx}=\dfrac{3d\left( x \right)}{dx}=3$
So, combining these we get,
$\dfrac{d\left[ {{\cos }^{2}}\left( 3x \right) \right]}{dx}=2\cos \left( 3x \right)\times \left( -\sin \left( 3x \right) \right)\times 3$
Simplifying we get,
$=-6\sin 3x\cos 3x$
We can further simplify it using the identity, $\sin 2x=2\sin x\cos x$, so we have,
$\begin{align}
  & 2\sin 3x\cos 3x=\sin \left( 2\times 3x \right) \\
 & \Rightarrow 2\sin 3x\cos 3x=\sin 6x \\
\end{align}$
So, our solution becomes, $-3\sin \left( 6x \right)$.

Note: While simplifying dfractions, we need to remember that we can add only the like terms. While calculating the derivative we need to be careful while picking the functions. If one function is comprised of more than one function then its derivative will be different than the usual, like $\dfrac{d\left( {{e}^{2x}} \right)}{dx}\ne {{e}^{2x}}$ as it is made up by the composition of ${{e}^{x}}$ and $2x$. Also, while differentiating we need to carefully use $d\left( {{x}^{n}} \right)=n.{{x}^{n-1}}$, simple errors may happen while calculating.