Question

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

Answer 1

Hint: The given function ${{\cos }^{2}}\left( 2x \right)$ is a composite function. We have to use the chain rule for the derivation. We derive the main function with respect to the secondary one. Then we take the derivation of the secondary function with respect to $x$. We take the multiplication of these two functions.

Complete step by step solution:
We differentiate the given function $f\left( x \right)={{\cos }^{2}}\left( 2x \right)$ with respect to $x$ using the chain rule.
Here we have a composite function where the main function is $g\left( x \right)={{\cos }^{2}}x$ and the other function is $h\left( x \right)=2x$.
We have $goh\left( x \right)=g\left( 2x \right)={{\cos }^{2}}\left( 2x \right)$. We take this as ours $f\left( x \right)={{\cos }^{2}}\left( 2x \right)$.
We need to find the value of $\dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{dx}\left[ {{\cos }^{2}}\left( 2x \right) \right]$. We know $f\left( x \right)=goh\left( x \right)$.
Differentiating $f\left( x \right)=goh\left( x \right)$, we get
\[\dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{dx}\left[ goh\left( x \right) \right]=\dfrac{d}{d\left[ h\left( x \right) \right]}\left[ goh\left( x \right) \right]\times \dfrac{d\left[ h\left( x \right) \right]}{dx}={{g}^{'}}\left[ h\left( x \right) \right]{{h}^{'}}\left( x \right)\].
The above-mentioned rule is the chain rule.
The chain rule allows us to differentiate with respect to the function $h\left( x \right)$ instead of $x$ and after that we need to take the differentiated form of $h\left( x \right)$ with respect to $x$.
For the function $f\left( x \right)={{\cos }^{2}}\left( 2x \right)$, we take differentiation of $f\left( x \right)={{\cos }^{2}}\left( 2x \right)$ with respect to the function $h\left( x \right)=2x$ instead of $x$ and after that we need to take the differentiated form of $h\left( x \right)=2x$ with respect to $x$.
We know the multiple angle formula of \[\sin 2x=2\sin x\cos x\].
The differentiation of $g\left( x \right)={{\cos }^{2}}x$ is ${{g}^{'}}\left( x \right)=2\cos x\dfrac{d\left( \cos x \right)}{dx}=-2\sin x\cos x=-\sin 2x$ and differentiation of $h\left( x \right)=2x$ is \[{{h}^{'}}\left( x \right)=2\]. We apply the formula of \[\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}\].
\[\Rightarrow \dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{d\left[ 2x \right]}\left[ {{\cos }^{2}}\left( 2x \right) \right]\times \dfrac{d\left[ 2x \right]}{dx}\]
We place the values of the differentiations and get
\[\Rightarrow \dfrac{d}{dx}\left[ f\left( x \right) \right]=\left( -\sin 4x \right)\left[ 2 \right]=-2\sin 4x\]
Therefore, the differentiation of ${{\cos }^{2}}\left( 2x \right)$ is \[-2\sin \left( 4x \right)\].

Note: We can also assume the secondary function as a new variable. For the given function ${{\cos }^{2}}\left( 2x \right)$, we assume $z=2x$. Then we use the chain rule in the form of
\[\dfrac{d}{dx}\left[ f\left( x \right) \right]=\dfrac{d}{dx}\left[ g\left( z \right) \right]=\dfrac{d}{dz}\left[ g\left( z \right) \right]\times \dfrac{dz}{dx}={{g}^{'}}\left( z \right)\times {{z}^{'}}\].
We replace the value $h\left( x \right)=z=2x$.