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 given function is: $$y = (\tan x)^x.$$ 
Take the natural logarithm on both sides to simplify the power: $$\ln y = x \ln (\tan x).$$ 

Differentiate both sides with respect to $x$: $$\frac{1}{y} \cdot \frac{dy}{dx} = \ln (\tan x) + x \cdot \frac{1}{\tan x} \cdot \sec^2 x.$$ 

Simplify: $$\frac{1}{y} \cdot \frac{dy}{dx} = \ln (\tan x) + x \cdot \frac{\sec^2 x}{\tan x}.$$ 

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

Simplify further: $$\frac{dy}{dx} = (\tan x)^x \left[\ln (\tan x) + x \cdot \frac{1}{\sin x \cos x}\right].$$ 

Final answer: $$\frac{dy}{dx} = (\tan x)^x \left[\ln (\tan x) + x \cdot \csc x \sec x\right].$$