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Yx+XX+Xy= ab

  Find dy/dx...?

 Jul 25, 2017
 #1
avatar+20805 
+1

Find dy/dx

Yx+XX+Xy= ab

 

\(\begin{array}{|rcll|} \hline y^x+x^x+x^y &=& a^b \\ y^x+x^x+x^y -a^b &=& 0 \\ \hline \end{array} \)

 

Formula:

\(\begin{array}{|rcll|} \hline \frac{dy}{dx} = - \dfrac{F_x}{F_y} \\ \hline \end{array} \)

 

\(\mathbf{F_x =\ ?} \begin{array}{|rcll|} \hline F_x &=& \frac{d }{dx}y^x + \frac{d }{dx}x^x + \frac{d }{dx}x^y + \frac{d }{dx}a^b \\\\ h &=& y^x \quad & | \quad \ln() \\ \ln(h) &=& x\cdot \ln(y) \\ \frac{h'}{h} &=& \ln(y) \\ h' &=& h\cdot \ln(y) \\ h' &=& y^x\cdot \ln(y) \\ \mathbf{ \frac{d }{dx}y^x } & \mathbf{=} & \mathbf{ y^x\cdot \ln(y) } \\\\ h &=& x^x \quad & | \quad \ln() \\ \ln(h) &=& x\cdot \ln(x) \\ \frac{h'}{h} &=& x\cdot \frac{1}{x} + 1\cdot \ln(x) \\ h' &=& h \cdot \Big( 1+\ln(x) \Big) \\ \mathbf{ \frac{d }{dx}x^x } & \mathbf{=} & \mathbf{ x^x \cdot \Big( 1+ \ln(x) \Big) } \\\\ h &=& x^y \quad & | \quad \ln() \\ \ln(h) &=& y\cdot \ln(x) \\ \frac{h'}{h} &=& \frac{y}{x} \\ h' &=& h \cdot \frac{y}{x} \\ h' &=& x^y \cdot \frac{y}{x} \\ \mathbf{ \frac{d }{dx}x^y } & \mathbf{=} & \mathbf{ x^y \cdot \frac{y}{x} } \\\\ \mathbf{ \frac{d }{dx}a^b } & \mathbf{=} & \mathbf{ 0 } \\ \hline \end{array} \)

 

\(\mathbf{F_y =\ ?} \begin{array}{|rcll|} \hline F_y &=& \frac{d }{dy}y^x + \frac{d }{dy}x^x + \frac{d }{dy}x^y +\frac{d }{dy}a^b \\\\ h &=& y^x \quad & | \quad \ln() \\ \ln(h) &=& x\cdot \ln(y) \\ \frac{h'}{h} &=& \frac{x}{y} \\ h' &=& h\cdot \frac{x}{y} \\ h' &=& y^x\cdot \frac{x}{y} \\ \mathbf{ \frac{d }{dy}y^x } & \mathbf{=} & \mathbf{ y^x\cdot \frac{x}{y} } \\\\ \mathbf{ \frac{d }{dy}x^x } & \mathbf{=} & \mathbf{ 0 } \\ h &=& x^y \quad & | \quad \ln() \\ \ln(h) &=& y\cdot \ln(x) \\ \frac{h'}{h} &=& \ln(x) \\ h' &=& h \cdot \ln(x) \\ h' &=& x^y \cdot \ln(x) \\ \mathbf{ \frac{d }{dy}x^y } & \mathbf{=} & \mathbf{ x^y \cdot \ln(x) } \\\\ \mathbf{ \frac{d }{dy}a^b } & \mathbf{=} & \mathbf{ 0 } \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \frac{dy}{dx} &=& - \dfrac{F_x}{F_y} \\\\ \frac{dy}{dx} &=& - \dfrac{ y^x\cdot \ln(y) + x^x \cdot \Big( 1+ \ln(x) \Big) + x^y \cdot \frac{y}{x} + 0 } { y^x\cdot \frac{x}{y} + 0 + x^y \cdot \ln(x) + 0 } \\\\ \frac{dy}{dx} &=& - \dfrac{ y^x\cdot \ln(y) + x^x \cdot \Big( 1+ \ln(x) \Big) + x^y \cdot \frac{y}{x} } { y^x\cdot \frac{x}{y} + x^y \cdot \ln(x) } \\\\ \frac{dy}{dx} &=& - \dfrac{ y^x\cdot \ln(y) + x^x \cdot \Big( 1+ \ln(x) \Big) + x^{y-1} \cdot y } { y^{x-1}\cdot x + x^y \cdot \ln(x) } \\\\ \mathbf{ \frac{dy}{dx} } & \mathbf{=} & \mathbf{ - \dfrac{ y^x\cdot \ln(y) + x^x \cdot \Big( 1+ \ln(x) \Big) + x^{y-1} \cdot y } { x^y \cdot \ln(x) + y^{x-1}\cdot x } } \\ \hline \end{array}\)

 

laugh

 Jul 25, 2017

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