Question

Differentiate: \[ y = (\tan x)^{\cot x} + (\sin x)^{\cos x}. \]

Hint:

For differentiation of power functions with variable bases and exponents, use logarithmic differentiation effectively.

Answer 1

For the first term:

\[ y_1 = (\tan x)^{\cot x}. \]

Taking the logarithm:

\[ \ln y_1 = \cot x \ln(\tan x). \]

Differentiating:

\[ \frac{1}{y_1} \frac{dy_1}{dx} = -\csc^2 x \ln(\tan x) + \cot x \cdot \frac{\sec^2 x}{\tan x}. \]

Thus:

\[ \frac{dy_1}{dx} = (\tan x)^{\cot x} \left(-\csc^2 x \ln(\tan x) + \frac{\cot x \sec^2 x}{\tan x} \right). \]

For the second term:

\[ y_2 = (\sin x)^{\cos x}. \]

Taking the logarithm:

\[ \ln y_2 = \cos x \ln(\sin x). \]

Differentiating:

\[ \frac{1}{y_2} \frac{dy_2}{dx} = -\sin x \ln(\sin x) + \cos x \cot x. \]

Thus:

\[ \frac{dy_2}{dx} = (\sin x)^{\cos x} \left(-\ln(\sin x) \sin x + \cos x \cot x \right). \]

Finally:

\[ \frac{dy}{dx} = \frac{dy_1}{dx} + \frac{dy_2}{dx}. \]