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, \[ \frac{d(2\cos^2 x)}{d(\cos^2 x)} = \mathbf{2} \]
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.
Let \( y = 2\cos^2 x \) and we wish to differentiate it with respect to \( z = \cos^2 x \), i.e.: \[ \frac{dy}{dz} = \frac{d}{d(\cos^2 x)}\left( 2\cos^2 x \right). \]
First, differentiate \( y = 2\cos^2 x \) with respect to \(x\): \[ \frac{dy}{dx} = \frac{d}{dx}\left( 2\cos^2 x \right) = 4\cos x (-\sin x) = -4\cos x \sin x. \]
\[ \frac{dz}{dx} = \frac{d}{dx}\left( \cos^2 x \right) = 2\cos x (-\sin x) = -2\cos x \sin x. \]
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: \[ \boxed{2}. \]
If $y = 5 \cos x - 3 \sin x$, prove that $\frac{d^2y}{dx^2} + y = 0$.
Flowering plants with hermaphrodite flowers have developed many reproductive strategies to ensure cross-pollination. Study the given outbreeding devices adopted by certain flowering plants and answer the questions that follow.
Note : All plants belong to the same species. No pollen tube growth/inhibition of pollen germination on stigma. Pollen germination on stigma.