Question

If \(y = (\tan x)^x\), then find \(\frac{dy}{dx}\).

Hint:

For functions of the form \(y = [f(x)]^{g(x)}\), take the natural logarithm on both sides to handle the variable exponent, and then differentiate using the product rule and chain rule.

Answer 1

The function to be differentiated is: \[ y = (\tan x)^x. \]
Taking the natural logarithm of both sides yields: \[ \ln y = x \ln (\tan x). \]

Differentiating both sides with respect to \(x\) results in: \[ \frac{1}{y} \cdot \frac{dy}{dx} = \ln (\tan x) + x \cdot \frac{1}{\tan x} \cdot \sec^2 x. \]

Simplifying the expression gives: \[ \frac{1}{y} \cdot \frac{dy}{dx} = \ln (\tan x) + x \cdot \frac{\sec^2 x}{\tan x}. \]

Multiplying by \(y = (\tan x)^x\) to isolate \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = (\tan x)^x \left[\ln (\tan x) + x \cdot \frac{\sec^2 x}{\tan x}\right]. \]

Further simplification leads to: \[ \frac{dy}{dx} = (\tan x)^x \left[\ln (\tan x) + x \cdot \frac{1}{\sin x \cos x}\right]. \]

The final derivative is: \[ \frac{dy}{dx} = (\tan x)^x \left[\ln (\tan x) + x \cdot \csc x \sec x\right]. \]