Question

How do you differentiate ${{\cos }^{2}}\left( {{x}^{2}} \right)$?

Answer 1

Hint: Now to differentiate the given function we will have to use chain rule of differentiation. Now first we will simplify the function ${{\cos }^{2}}\left( {{x}^{2}} \right)$ by using the identity ${{\cos }^{2}}x=\dfrac{1+\cos 2x}{2}$ . Now we will differentiate the function using chain rule of differentiation which states $\dfrac{d\left( f\left( g\left( x \right) \right) \right)}{dx}=f'\left( g\left( x \right) \right).g'\left( x \right)$ . Now we know that differentiation of $\cos x$ is $-\sin x$ and differentiation of ${{x}^{n}}$ is $n{{x}^{n-1}}$ . Hence we will use these values to find the differentiation of the function.

Complete step-by-step answer:
Now the given function ${{\cos }^{2}}\left( {{x}^{2}} \right)$ is a composite function of the form $f\left( g\left( x \right) \right)$ .
To differentiate the function we know we have to use chain rule of differentiation. Now we have no standard differentiation for the function ${{\cos }^{2}}x$ . Hence first we will use trigonometric identities to simplify the function ${{\cos }^{2}}x$ such that the obtained function can be easily differentiable.
Now we know that ${{\cos }^{2}}x=\dfrac{\cos \left( 2x \right)+1}{2}$ .
Hence we have ${{\cos }^{2}}{{x}^{2}}=\dfrac{1+\cos 2{{x}^{2}}}{2}$
Now let us differentiate $\dfrac{1+\cos 2{{x}^{2}}}{2}$
Now according to chain rule we have $f\left( x \right)=\dfrac{1+\cos x}{2}$ and $g\left( x \right)=2{{x}^{2}}$
Now we that $f'\left( cx \right)=cf'\left( x \right)$ we can take $\dfrac{1}{2}$ out of denominator Hence we have $f'\left( x \right)=\dfrac{-\sin x}{2}$.
 Also since we know $\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}}$ We have can write $g'\left( x \right)=4x$
Now hence we have $f'\left( g\left( x \right) \right)=\dfrac{-\sin \left( 2{{x}^{2}} \right)}{2}$ and $g'\left( x \right)=4x$
Now we know that according to chain rule we have differentiation of the function $f\left( g\left( x \right) \right)=f'\left( g\left( x \right) \right)g'\left( x \right)$
Now using chain rule we have differentiation of $\dfrac{1+\cos 2{{x}^{2}}}{2}$ is $\dfrac{-\sin 2{{x}^{2}}\left( 4x \right)}{2}$
But since we have ${{\cos }^{2}}x=\dfrac{\cos \left( 2x \right)+1}{2}$ we can say that the differentiation of ${{\cos }^{2}}x$ is $-\sin \left( 2{{x}^{2}} \right)\left( 2x \right)$ .
Hence the given function is differentiated.

Note: Now note that while differentiation a composite function with the help of chain rule we have $f'\left( g\left( x \right) \right).g'\left( x \right)$ and not $f'\left( x \right).g'\left( x \right)$ . Hence remember to find the value of $f'\left( g\left( x \right) \right)$ after differentiating the function $f\left( x \right)$ . Also not to be confused between $f\left( x \right).g\left( x \right)$ and $f\left( g\left( x \right) \right)$ .