Appeared in: 2024 (5 sets)
Derivative of $e^{\sin^2 x}$ with respect to $\cos x$ is : (A) $\sin x \, e^{\sin^2 x}$ (B) $\cos x \, e^{\sin^2 x}$ (C) $-2 \cos x \, e^{\sin^2 x}$ (D) $-2 \sin^2 x \cos x \, e^{\sin^2 x}$
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Official Evaluation Steps
Step 1 (Formula / Governing Law):
To find the derivative of a function $u(x)$ with respect to another function $v(x)$, we use the chain rule formula:
$$\frac{du}{dv} = \frac{du/dx}{dv/dx}$$where $u = e^{\sin^2 x}$ and $v = \cos x$. (0.25 Mark)
Step 2 (Step-by-Step Working):
- Differentiating $u = e^{\sin^2 x}$ with respect to $x$: $$\frac{du}{dx} = e^{\sin^2 x} \cdot \frac{d}{dx}(\sin^2 x) = e^{\sin^2 x} \cdot (2 \sin x \cos x)$$
- Differentiating $v = \cos x$ with respect to $x$: $$\frac{dv}{dx} = -\sin x$$
- Applying the quotient of derivatives: $$\frac{du}{dv} = \frac{e^{\sin^2 x} \cdot 2 \sin x \cos x}{-\sin x}$$
- Simplifying the expression (canceling $\sin x$): $$\frac{du}{dv} = -2 \cos x \cdot e^{\sin^2 x}$$
Step 3 (Final Answer):
Comparing the result with the given options, the correct choice is (C).
$$\boxed{-2 \cos x \, e^{\sin^2 x}}$$ (0.25 Mark)