Question

if \(2x^y+3y^x=20\) then \(\frac{dy}{dx}\) at (2,2) is equal to:

• A. \(-(\frac{3+log_e4}{ 2+loge_8})\)
• B. \(-(\frac{3+log_e16}{ 4+loge_8})\)
• C. \(-(\frac{3+log_e8}{ 2+loge_4 })\)
• D. \(-(\frac{2+log_e8}{ 3+loge_4})\)

Answer 1

The Correct Option is D

To find \(\frac{dy}{dx}\) for the equation \(2x^y + 3y^x = 20\) at the point (2,2), we must differentiate implicitly with respect to \(x\).

1. Calculate the derivative of \(2x^y\):
   • Differentiate using the chain rule: \(\frac{d}{dx}(x^y) = yx^{y-1} + x^y \ln(x) \cdot \frac{dy}{dx}\)
   • Thus, \(\frac{d}{dx}(2x^y) = 2\left(yx^{y-1} + x^y \ln(x) \cdot \frac{dy}{dx}\right)\)
2. Calculate the derivative of \(3y^x\):
   • Again, use the chain rule: \(\frac{d}{dx}(y^x) = y^x \ln(y) + xy^{x-1} \cdot \frac{dy}{dx}\)
   • Thus, \(\frac{d}{dx}(3y^x) = 3\left( y^x \ln(y) + xy^{x-1} \cdot \frac{dy}{dx} \right)\)
3. Differentiate the entire equation with respect to \(x\): \[\frac{d}{dx}(2x^y + 3y^x) = \frac{d}{dx}(20)\]
   • Combine the derivatives: \(2\left(yx^{y-1} + x^y \ln(x) \cdot \frac{dy}{dx}\right) + 3\left( y^x \ln(y) + xy^{x-1} \cdot \frac{dy}{dx} \right) = 0\)
4. Substitute \(x=2, y=2\) into the differentiated equation:
   • For \(2x^y\): 2\left(2 \cdot 2^{2-1} + 2^2 \ln(2) \cdot \frac{dy}{dx}\right)\)
   • = \(2(4+4 \ln(2) \cdot \frac{dy}{dx})\) = \((8+8 \ln(2) \cdot \frac{dy}{dx})\)
   • For \(3y^x\): 3\left( 2^2 \ln(2) + 2 \cdot 2^{2-1} \cdot \frac{dy}{dx} \right)\)
   • = \((12 \ln(2) + 12 \cdot \frac{dy}{dx})\)
5. Solve the equation: \[(8 + 8 \ln(2) \cdot \frac{dy}{dx}) + (12 \ln(2) + 12 \cdot \frac{dy}{dx}) = 0\]
   • Combine terms: \(8 + 12 \ln(2) + (8 \ln(2) + 12) \cdot \frac{dy}{dx} = 0\)
   • Isolate \(\frac{dy}{dx}\):
   • \(\frac{dy}{dx} = -\frac{8 + 12 \ln(2)}{8 \ln(2) + 12}\)
   • Simplify the expression using \(\ln(8) = 3\ln(2)\): \(-\frac{2 + \ln(8)}{3 + \ln(4)}\)

The correct answer is -\left(\frac{2 + \ln_e 8}{3 + \ln_e 4}\right).