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

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

Explanation

Solution

We are given a trigonometric function, the cosine function which is in turn a function of x variable. In order to simplify this equation, we must do a little manipulation to apply basic trigonometric identities which has been derived and proven by application on the Pythagorean triangle. Then, we will differentiate the equation and bring it to its simplest form.

Complete step-by-step solution:
We must have a prior knowledge of basic trigonometric identities to solve this particular problem. We are given the square of the cosine function. We shall add and subtract 1 from the equation to bring it to the required form. Hence, we shall use the half angle identity,
2cos⁡2θ=cos⁡2θ+12{{\cos }^{2}}\theta =\cos 2\theta +1
⇒2cos⁡2θ−1=cos⁡2θ\Rightarrow 2{{\cos }^{2}}\theta -1=\cos 2\theta
Adding and subtracting 1 from 2(cos⁡2x)2\left( {{\cos }^{2}}x \right), we get
⇒2(cos⁡2x)=2(cos⁡2x)−1+1\Rightarrow 2\left( {{\cos }^{2}}x \right)=2\left( {{\cos }^{2}}x \right)-1+1
In our given equation, θ=x\theta =x, on applying the identity on the given equation, (2cos⁡2x−1)\left( 2{{\cos }^{2}}x-1 \right), we get:
⇒2(cos⁡2x)=cos⁡2x+1\Rightarrow 2\left( {{\cos }^{2}}x \right)=\cos 2x+1
Differentiating this equation with respect to x, we get
⇒ddx2(cos⁡2x)=ddx(cos⁡2x+1) ⇒ddx2(cos⁡2x)=ddxcos⁡2x+ddx1 \begin{aligned} & \Rightarrow \dfrac{d}{dx}2\left( {{\cos }^{2}}x \right)=\dfrac{d}{dx}\left( \cos 2x+1 \right) \\\ & \Rightarrow \dfrac{d}{dx}2\left( {{\cos }^{2}}x \right)=\dfrac{d}{dx}\cos 2x+\dfrac{d}{dx}1 \\\ \end{aligned}
We know by properties of basic differentiation that ddx(cos⁡ax)=−asin⁡ax\dfrac{d}{dx}\left( \cos ax \right)=-a\sin ax and the derivative of a constant term is equal to zero. Thus, applying this property, we get
⇒ddx2(cos⁡2x)=−2sin⁡2x+0\Rightarrow \dfrac{d}{dx}2\left( {{\cos }^{2}}x \right)=-2\sin 2x+0
∴ddx2(cos⁡2x)=−2sin⁡2x\therefore \dfrac{d}{dx}2\left( {{\cos }^{2}}x \right)=-2\sin 2x
Therefore, the derivative of 2(cos⁡2x)2\left( {{\cos }^{2}}x \right) is given as −2sin⁡2x-2\sin 2x.

Note: We must try to make the whole trigonometric equation with respect to any one trigonometric function only. While simplifying any trigonometric equation, we must always try to simplify them as the sine or cosine functions. This is because only these two functions are the simplest of all the trigonometric functions derived directly from the Pythagorean theory.