Solveeit Logo
Login

Question

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

How do you differentiate cos⁡2(x2){{\cos }^{2}}\left( {{x}^{2}} \right)?

Explanation

Solution

Now to differentiate the given function we will have to use chain rule of differentiation. Now first we will simplify the function cos⁡2(x2){{\cos }^{2}}\left( {{x}^{2}} \right) by using the identity cos⁡2x=1+cos⁡2x2{{\cos }^{2}}x=\dfrac{1+\cos 2x}{2} . Now we will differentiate the function using chain rule of differentiation which states d(f(g(x)))dx=f′(g(x)).g′(x)\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\cos x is −sin⁡x-\sin x and differentiation of xn{{x}^{n}} is nxn−1n{{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(x2){{\cos }^{2}}\left( {{x}^{2}} \right) is a composite function of the form f(g(x))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⁡2x{{\cos }^{2}}x . Hence first we will use trigonometric identities to simplify the function cos⁡2x{{\cos }^{2}}x such that the obtained function can be easily differentiable.
Now we know that cos⁡2x=cos⁡(2x)+12{{\cos }^{2}}x=\dfrac{\cos \left( 2x \right)+1}{2} .
Hence we have cos⁡2x2=1+cos⁡2x22{{\cos }^{2}}{{x}^{2}}=\dfrac{1+\cos 2{{x}^{2}}}{2}
Now let us differentiate 1+cos⁡2x22\dfrac{1+\cos 2{{x}^{2}}}{2}
Now according to chain rule we have f(x)=1+cos⁡x2f\left( x \right)=\dfrac{1+\cos x}{2} and g(x)=2x2g\left( x \right)=2{{x}^{2}}
Now we that f′(cx)=cf′(x)f'\left( cx \right)=cf'\left( x \right) we can take 12\dfrac{1}{2} out of denominator Hence we have f′(x)=−sin⁡x2f'\left( x \right)=\dfrac{-\sin x}{2}.
Also since we know dxndx=nxn−1\dfrac{d{{x}^{n}}}{dx}=n{{x}^{n-1}} We have can write g′(x)=4xg'\left( x \right)=4x
Now hence we have f′(g(x))=−sin⁡(2x2)2f'\left( g\left( x \right) \right)=\dfrac{-\sin \left( 2{{x}^{2}} \right)}{2} and g′(x)=4xg'\left( x \right)=4x
Now we know that according to chain rule we have differentiation of the function f(g(x))=f′(g(x))g′(x)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 1+cos⁡2x22\dfrac{1+\cos 2{{x}^{2}}}{2} is −sin⁡2x2(4x)2\dfrac{-\sin 2{{x}^{2}}\left( 4x \right)}{2}
But since we have cos⁡2x=cos⁡(2x)+12{{\cos }^{2}}x=\dfrac{\cos \left( 2x \right)+1}{2} we can say that the differentiation of cos⁡2x{{\cos }^{2}}x is −sin⁡(2x2)(2x)-\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′(g(x)).g′(x)f'\left( g\left( x \right) \right).g'\left( x \right) and not f′(x).g′(x)f'\left( x \right).g'\left( x \right) . Hence remember to find the value of f′(g(x))f'\left( g\left( x \right) \right) after differentiating the function f(x)f\left( x \right) . Also not to be confused between f(x).g(x)f\left( x \right).g\left( x \right) and f(g(x))f\left( g\left( x \right) \right) .