What is the derivative of $\log(\sin^2 x)$ with respect to $\sin x$?
Show 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.
Step 1 : Understanding the Question:
The question asks us to compute the derivative of the function $f(x) = \log(\sin^2 x)$ with respect to another function $g(x) = \sin x$. Step 2 : Key Formulas and Approach:
For finding the derivative of a function $u$ with respect to another variable $v$, we use the change of variable method.
Let $u = \sin x$. We can express $\log(\sin^2 x)$ in terms of $u$ and then differentiate directly with respect to $u$:
\[ \frac{d}{du}[\log(u^2)] \] Step 3 : Detailed Explanation:
Let us define the variable:
\[ u = \sin x \]
We want to find:
\[ \frac{d}{du} \left[ \log(\sin^2 x) \right] \]
Substitute $\sin x = u$ into the expression:
\[ \log(\sin^2 x) = \log(u^2) \]
Using the logarithmic power property $\log(a^b) = b \log a$:
\[ \log(u^2) = 2 \log u \]
Now, differentiate $2 \log u$ with respect to $u$:
\[ \frac{d}{du}(2 \log u) = \frac{2}{u} \]
Now substitute $u = \sin x$ back into the result:
\[ \frac{2}{\sin x} = 2 \csc x \] Step 4 : Final Answer:
The derivative with respect to $\sin x$ is $2\csc x$.
This corresponds to Option (A).
