Question

Find the differential of \( x^{\cot x} + \frac{2x^2-3}{2x^2-x+2} \) with respect to \( x \).

Hint:

For \( x^{f(x)} \), the derivative always takes the form \( x^{f(x)} [ \frac{f(x)}{x} + f'(x)\log x ] \). Applying this directly can save time.

Answer 1

Concept:
• Differentiate the two terms separately and then sum their derivatives.
• Use logarithmic differentiation for the function-to-power-function term \( x^{\cot x} \).
• Use the quotient rule for the rational algebraic function.

Step 1: Differentiate \( u = x^{\cot x} \)
Let \( u = x^{\cot x} \). Taking natural log on both sides:
\[ \log u = \cot x \log x \]
Differentiating both sides with respect to \( x \):
\[ \frac{1}{u} \frac{du}{dx} = (-\csc^2 x) \log x + \cot x \left( \frac{1}{x} \right) \]
\[ \frac{du}{dx} = x^{\cot x} \left( \frac{\cot x}{x} - \csc^2 x \log x \right) \]

Step 2: Differentiate \( v = \frac{2x^2-3}{2x^2-x+2} \)
Using the quotient rule \( \frac{d}{dx}(\frac{N}{D}) = \frac{N'D - ND'}{D^2} \):
\[ \frac{dv}{dx} = \frac{(4x)(2x^2-x+2) - (2x^2-3)(4x-1)}{(2x^2-x+2)^2} \]
Expanding the numerator:
\[ = \frac{(8x^3 - 4x^2 + 8x) - (8x^3 - 2x^2 - 12x + 3)}{(2x^2-x+2)^2} \]
\[ = \frac{-2x^2 + 20x - 3}{(2x^2-x+2)^2} \]

Step 3: Combine the results
Let \( y = u + v \). Then \( \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \).
\[ \frac{dy}{dx} = x^{\cot x} \left( \frac{\cot x}{x} - \csc^2 x \log x \right) + \frac{-2x^2 + 20x - 3}{(2x^2-x+2)^2} \]