Question

What is the derivative of $\log(\sin^2 x)$ with respect to $\sin x$?

• A. $2\csc x$
• B. $\sin 2x$
• C. $4\csc x$
• D. $\cot x \csc 2x$

Hint:

For any derivative of $f(g(x))$ with respect to $g(x)$, you can treat $g(x)$ as a single variable $u$.
This simplifies the problem into a simple single-variable derivative $\frac{d}{du}[f(u)]$, completely avoiding the need for complex chain-rule steps.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
To find the derivative of \(u\) with respect to \(v\), we use the formula \(\frac{du}{dv} = \frac{du/dx}{dv/dx}\).
Step 2: Differentiating the Functions:
Let \(u = \log(\sin^2 x) = 2 \log(\sin x)\).
\(\frac{du}{dx} = 2 \cdot \frac{1}{\sin x} \cdot \cos x = 2 \cot x\).
Let \(v = \sin x\).
\(\frac{dv}{dx} = \cos x\).
Step 3: Calculating the Ratio:
\(\frac{du}{dv} = \frac{2 \cot x}{\cos x} = \frac{2 (\cos x / \sin x)}{\cos x} = \frac{2}{\sin x} = 2 \csc x\).
Step 4: Final Answer:
The derivative is 2 cosec x.