Question:

Differentiate $2\cos^2 x$ w.r.t. $\cos^2 x$.

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To differentiate $\cos^2 x$, remember to use the chain rule as $\frac{d}{dx}[\cos^2 x] = 2\cos x \cdot \frac{d}{dx}[\cos x]$.
Updated On: Aug 28, 2025
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Approach Solution - 1

Given: Differentiate \( 2\cos^2 x \) with respect to \( \cos^2 x \). We can treat \( \cos^2 x \) as a single variable, say \( u \). Let \[ u = \cos^2 x \] Then the given expression becomes: \[ 2\cos^2 x = 2u \] Now, differentiating \( 2u \) with respect to \( u \): \[ \frac{d(2u)}{du} = 2 \] Therefore, \[ \frac{d(2\cos^2 x)}{d(\cos^2 x)} = \mathbf{2} \] 

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Approach Solution -2

We are asked to differentiate \(2\cos^2 x\) with respect to \(\cos^2 x\), which means finding the derivative of \(2\cos^2 x\) while treating \(\cos^2 x\) as the independent variable. 

Step 1: Rewrite the given function

Let \( y = 2\cos^2 x \) and we wish to differentiate it with respect to \( z = \cos^2 x \), i.e.: \[ \frac{dy}{dz} = \frac{d}{d(\cos^2 x)}\left( 2\cos^2 x \right). \]

Step 2: Apply the chain rule

First, differentiate \( y = 2\cos^2 x \) with respect to \(x\): \[ \frac{dy}{dx} = \frac{d}{dx}\left( 2\cos^2 x \right) = 4\cos x (-\sin x) = -4\cos x \sin x. \]

Step 3: Differentiate \( z = \cos^2 x \) with respect to \( x \)

\[ \frac{dz}{dx} = \frac{d}{dx}\left( \cos^2 x \right) = 2\cos x (-\sin x) = -2\cos x \sin x. \]

Step 4: Use the chain rule for \( \frac{dy}{dz} \)

By the chain rule: \[ \frac{dy}{dz} = \frac{\frac{dy}{dx}}{\frac{dz}{dx}} = \frac{-4\cos x \sin x}{-2\cos x \sin x} = 2. \]

Final Answer: \[ \boxed{2}. \]

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