Differentiate the function with respect to x: $\cos 4x\cos 2x$?

Hint: Assume the given function as y. Now, divide and multiply the given function with 2 and use the trigonometric identity $2\cos a\cos b=\cos \left( \dfrac{a+b}{2} \right)+\cos \left( \dfrac{a-b}{2} \right)$ to convert the product of the cosine functions into the sum. Differentiate both the sides with respect to x and use the formula $\dfrac{d\left( \cos \left( ax+b \right) \right)}{dx}=-a\sin \left( ax+b \right)$ to get the answer. Here, a and b are constants.

Complete step by step solution:
Here we have been provided with the function $\cos 4x\cos 2x$and we are asked to differentiate it. Let us assume the given function as y. So we have,
\[\Rightarrow y=\cos 4x\cos 2x\]
Multiplying the given expression with 2 and then to balance dividing it with 2 we get,
\[\Rightarrow y=\dfrac{1}{2}\left( 2\cos 4x\cos 2x \right)\]
Using the trigonometric identity $2\cos a\cos b=\cos \left( \dfrac{a+b}{2} \right)+\cos \left( \dfrac{a-b}{2} \right)$ we get,
\[\begin{align}
  & \Rightarrow y=\dfrac{1}{2}\left[ \cos \left( \dfrac{4x+2x}{2} \right)+\cos \left( \dfrac{4x-2x}{2} \right) \right] \\
 & \Rightarrow y=\dfrac{1}{2}\left[ \cos \left( 3x \right)+\cos \left( x \right) \right] \\
\end{align}\]
Differentiating both the sides with respect to x we get,
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( \dfrac{1}{2}\left[ \cos \left( 3x \right)+\cos \left( x \right) \right] \right)}{dx}\]
Since $\dfrac{1}{2}$ is a constant so we can take it out from the derivative, therefore we get,
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2}\times \dfrac{d\left( \left[ \cos \left( 3x \right)+\cos \left( x \right) \right] \right)}{dx}\]
Breaking the terms of the derivative and using the formula $\dfrac{d\left( \cos \left( ax+b \right) \right)}{dx}=-a\sin \left( ax+b \right)$, where a and b are constants, we get,
\[\begin{align}
  & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2}\times \left[ \dfrac{d\left( \cos \left( 3x \right) \right)}{dx}+\dfrac{d\left( \cos \left( x \right) \right)}{dx} \right] \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2}\times \left[ -3\sin 3x-\sin x \right] \\
 & \therefore \dfrac{dy}{dx}=\dfrac{-1}{2}\times \left[ 3\sin 3x+\sin x \right] \\
\end{align}\]
Hence, the above relation is our answer.

Note: You must remember all the basic rules and formulas of differentiation like: - product rule, chain rule, \[\dfrac{u}{v}\] rule etc. Remember the derivative formulas of the functions like exponential, logarithmic, trigonometric functions etc. Note that the formula $\dfrac{d\left( \cos \left( ax+b \right) \right)}{dx}=-a\sin \left( ax+b \right)$ is a result of chain rule of derivative. You can also solve the question using the product rule of derivative where you have to assume $u=\cos 4x$ and $v=\cos 2x$ and apply the formula $\dfrac{d\left( u\times v \right)}{dx}=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$. The final answer may seem different but on simplification using different trigonometric identities we will get the same answer.