Step 1 : Understanding the Question:
The topic for this question is Differentiation, specifically using the Power of a Power rule and Logarithmic Differentiation. The function is a "nested" exponent where $x$ is raised to the power of $\sin x$, and that entire result is raised to the power of $\tan x$. This can be simplified into a single exponent first, and then differentiated using the natural logarithm to handle the variable in the exponent.
Step 2 : Key Formulas and approach:
1. Power Rule of Exponents: $(a^m)^n = a^{m \cdot n}$.
2. Logarithmic Differentiation: If $\ln y = f(x)$, then $\frac{1}{y} y' = f'(x)$.
3. Product Rule for Three Functions: $(uvw)' = u'vw + uv'w + uvw'$.
4. Approach: Simplify the expression to $y = x^{\sin x \tan x}$, take $\ln$ of both sides, differentiate the product on the right side, and solve for $dy/dx$.
Step 3 : Detailed Explanation:
Step 1: Simplify the given expression using the rule $(a^m)^n = a^{mn}$. We get $y = x^{\sin x \tan x}$.
Step 2: Take the natural logarithm on both sides: $\ln y = \ln (x^{\sin x \tan x}) = \sin x \cdot \tan x \cdot \ln x$.
Step 3: Differentiate both sides with respect to $x$. On the left, we get $\frac{1}{y} \frac{dy}{dx}$.
Step 4: Apply the product rule to the right side $\frac{d}{dx} (\sin x \cdot \tan x \cdot \ln x)$. We have three factors: $u = \sin x$, $v = \tan x$, and $w = \ln x$.
Term 1 ($u'vw$): $\cos x \cdot \tan x \cdot \ln x$.
Term 2 ($uv'w$): $\sin x \cdot \sec^2 x \cdot \ln x$.
Term 3 ($uvw'$): $\sin x \cdot \tan x \cdot \frac{1}{x}$.
Step 5: Sum these terms: $\frac{1}{y} \frac{dy}{dx} = \sec^2 x \sin x \ln x + \tan x \cos x \ln x + \frac{\tan x \sin x}{x}$.
Step 6: Multiply by $y$ to get the final derivative: $\frac{dy}{dx} = y \left[\sec^2 x \sin x \ln x + \tan x \cos x \ln x + \frac{\tan x \sin x}{x}\right]$.
Step 4 : Final Answer:
By simplifying the exponent and applying logarithmic differentiation with the product rule, the derivative matches Option (A).