Question

Find the derivative of the composite algebraic-trigonometric function expression with respect to $x$: \[ f(x) = x^{\cot x} + \frac{2x^2 - 3}{2x^2 - x + 2} \]

Hint:

When applying the quotient rule, be extra careful with negative signs when expanding polynomial terms in the numerator. A single sign error during expansion will change the final quadratic coefficients!

Answer 1

Concept: The given function is a sum of two distinct mathematical terms: a variable base raised to a variable exponent ($u = x^{\cot x}$), and a rational algebraic fraction ($v = \frac{2x^2 - 3}{2x^2 - x + 2}$). We compute their derivatives independently and then sum them together: \[ \frac{df}{dx} = \frac{du}{dx} + \frac{dv}{dx} \] We use logarithmic differentiation to evaluate $\frac{du}{dx}$, and the quotient rule ($\frac{d}{dx}\left[\frac{p}{q}\right] = \frac{p'q - pq'}{q^2}$) to evaluate $\frac{dv}{dx}$.

Step 1: Differentiating the variable exponent component $u = x^{\cot x}$ using logarithms.
Let: \[ u = x^{\cot x} \] Taking the natural logarithm on both sides: \[ \log u = \cot x \cdot \log x \] Differentiating both sides with respect to $x$ using the product rule: \[ \frac{1}{u}\frac{du}{dx} = \frac{d}{dx}(\cot x) \cdot \log x + \cot x \cdot \frac{d}{dx}(\log x) \] We know that $\frac{d}{dx}(\cot x) = -\csc^2 x$ and $\frac{d}{dx}(\log x) = \frac{1}{x}$: \[ \frac{1}{u}\frac{du}{dx} = -\csc^2 x \cdot \log x + \frac{\cot x}{x} \] Multiply by $u$ to isolate $\frac{du}{dx}$, then substitute back $u = x^{\cot x}$: \[ \frac{du}{dx} = x^{\cot x} \left[ \frac{\cot x}{x} - \csc^2 x \log x \right] \quad \cdots (1) \]

Step 2: Differentiating the rational expression $v = \frac{2x^2 - 3}{2x^2 - x + 2}$ using the quotient rule.
Let: \[ v = \frac{2x^2 - 3}{2x^2 - x + 2} \] Apply the quotient rule formula: \[ \frac{dv}{dx} = \frac{(4x)(2x^2 - x + 2) - (2x^2 - 3)(4x - 1)}{(2x^2 - x + 2)^2} \] Let us expand both polynomial products in the numerator:
• First product: $4x(2x^2 - x + 2) = 8x^3 - 4x^2 + 8x$
• Second product: $(2x^2 - 3)(4x - 1) = 8x^3 - 2x^2 - 12x + 3$ Subtract the second product from the first: \[ \text{Numerator} = (8x^3 - 4x^2 + 8x) - (8x^3 - 2x^2 - 12x + 3) \] \[ = 8x^3 - 4x^2 + 8x - 8x^3 + 2x^2 + 12x - 3 \] Group and combine like terms: \[ = (-4x^2 + 2x^2) + (8x + 12x) - 3 = -2x^2 + 20x - 3 \] Thus, the derivative of the second component is: \[ \frac{dv}{dx} = \frac{-2x^2 + 20x - 3}{(2x^2 - x + 2)^2} \quad \cdots (2) \]

Step 3: Combining the results into the final derivative expression.
Sum the two independent derivatives from equations (1) and (2): \[ \frac{df}{dx} = x^{\cot x} \left[ \frac{\cot x}{x} - \csc^2 x \log x \right] + \frac{-2x^2 + 20x - 3}{(2x^2 - x + 2)^2} \] This expression represents the complete derivative of the original function.