if \(2x^y+3y^x=20\) then \(\frac{dy}{dx}\) at (2,2) is equal to:
- \(-(\frac{3+log_e4}{ 2+loge_8})\)
- \(-(\frac{3+log_e16}{ 4+loge_8})\)
- \(-(\frac{3+log_e8}{ 2+loge_4 })\)
- \(-(\frac{2+log_e8}{ 3+loge_4})\)
The Correct Option is D
Solution and Explanation
Given the equation: \(2x^y + 3y^x = 20\). We need to find \(\frac{dy}{dx}\) at \((2,2)\). Differentiating both sides with respect to \(x\): \[ 2 \frac{d}{dx}(x^y) + 3 \frac{d}{dx}(y^x) = 0. \] Using the product rule and chain rule: \[ 2x^y \left( \frac{y}{x} + \ln(x) \frac{dy}{dx} \right) + 3y^x \left( \frac{x}{y} \frac{dy}{dx} + \ln(y) \right) = 0. \] Substituting \(x = 2\) and \(y = 2\): \[ 8 \left( \frac{1}{2} + \ln(2) \frac{dy}{dx} \right) + 12 \left( \frac{1}{2} \frac{dy}{dx} + \ln(2) \right) = 0. \] Simplify and solve for \(\frac{dy}{dx}\): \[ \frac{dy}{dx} = - \frac{2 + \ln(8)}{3 + \ln(4)}. \] ✅ Therefore, \(\boldsymbol{\frac{dy}{dx} = - \frac{2 + \ln(8)}{3 + \ln(4)}}\).
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