Question:
Find the differential of \( x^{\cot x} + \frac{2x^2-3}{2x^2-x+2} \) with respect to \( x \).
Find the differential of \( x^{\cot x} + \frac{2x^2-3}{2x^2-x+2} \) with respect to \( x \).
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Break down functions with different rules (power vs fraction) into separate variables \( u \) and \( v \) to keep work organized.
Simplify numerator algebra step-by-step to avoid missing negative signs.
Simplify numerator algebra step-by-step to avoid missing negative signs.
Updated On: Sep 10, 2026
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Solution and Explanation
Concept:
• Sum rule for differentiation.
• Logarithmic differentiation for \( u(x)^{v(x)} \).
• Quotient rule: \( \frac{d}{dx}(\frac{u}{v}) = \frac{vu' - uv'}{v^2} \).
Step 1: Differentiate the first term \( u = x^{\cot x} \)
Let \( u = x^{\cot x} \). Taking natural log: \[ \log u = \cot x \log x \] Differentiating wrt \( x \): \[ \frac{1}{u}\frac{du}{dx} = -\text{cosec}^2 x \cdot \log x + \cot x \cdot \frac{1}{x} \] \[ \frac{du}{dx} = x^{\cot x} \left( \frac{\cot x}{x} - \log x \cdot \text{cosec}^2 x \right) \]
Step 2: Differentiate the second term \( v = \frac{2x^2-3}{2x^2-x+2} \)
Using quotient rule: \[ \frac{dv}{dx} = \frac{(2x^2-x+2)(4x) - (2x^2-3)(4x-1)}{(2x^2-x+2)^2} \] Numerator: \[ (8x^3 - 4x^2 + 8x) - (8x^3 - 2x^2 - 12x + 3) \] \[ 8x^3 - 4x^2 + 8x - 8x^3 + 2x^2 + 12x - 3 = -2x^2 + 20x - 3 \] \[ \frac{dv}{dx} = \frac{-2x^2 + 20x - 3}{(2x^2-x+2)^2} \]
Step 3: Combine the derivatives
The total derivative \( \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \): \[ \frac{dy}{dx} = x^{\cot x} \left( \frac{\cot x}{x} - \text{cosec}^2 x \log x \right) + \frac{20x - 2x^2 - 3}{(2x^2-x+2)^2} \]
• Sum rule for differentiation.
• Logarithmic differentiation for \( u(x)^{v(x)} \).
• Quotient rule: \( \frac{d}{dx}(\frac{u}{v}) = \frac{vu' - uv'}{v^2} \).
Step 1: Differentiate the first term \( u = x^{\cot x} \)
Let \( u = x^{\cot x} \). Taking natural log: \[ \log u = \cot x \log x \] Differentiating wrt \( x \): \[ \frac{1}{u}\frac{du}{dx} = -\text{cosec}^2 x \cdot \log x + \cot x \cdot \frac{1}{x} \] \[ \frac{du}{dx} = x^{\cot x} \left( \frac{\cot x}{x} - \log x \cdot \text{cosec}^2 x \right) \]
Step 2: Differentiate the second term \( v = \frac{2x^2-3}{2x^2-x+2} \)
Using quotient rule: \[ \frac{dv}{dx} = \frac{(2x^2-x+2)(4x) - (2x^2-3)(4x-1)}{(2x^2-x+2)^2} \] Numerator: \[ (8x^3 - 4x^2 + 8x) - (8x^3 - 2x^2 - 12x + 3) \] \[ 8x^3 - 4x^2 + 8x - 8x^3 + 2x^2 + 12x - 3 = -2x^2 + 20x - 3 \] \[ \frac{dv}{dx} = \frac{-2x^2 + 20x - 3}{(2x^2-x+2)^2} \]
Step 3: Combine the derivatives
The total derivative \( \frac{dy}{dx} = \frac{du}{dx} + \frac{dv}{dx} \): \[ \frac{dy}{dx} = x^{\cot x} \left( \frac{\cot x}{x} - \text{cosec}^2 x \log x \right) + \frac{20x - 2x^2 - 3}{(2x^2-x+2)^2} \]
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