How to differentiate e^(xy)+1/y=5x+6?

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Answer 1 · 2018-03-25T19:46:38

$\frac{d y}{d x} = \frac{5 - y e^{x y}}{x e^{x y} - \frac{1}{y^{2}}}$ $dy/dx=(5-ye^(xy))/(xe^(xy)-1/y^2)$

$e^{x y} + y^{- 1} = 5 x + 6$ $e^(xy)+y^-1=5x+6$

$e^{x y} \cdot \frac{d}{d x} (x y) - y^{- 2} \cdot \frac{d}{d x} (y) = 5$ $e^(xy)*d/dx(xy) -y^-2*d/dx(y)=5$

$e^{x y} (y + x \frac{d y}{d x}) - y^{- 2} \frac{d y}{d x} = 5$ $e^(xy)(y+xdy/dx)-y^-2dy/dx=5$

$y e^{x y} + x e^{x y} \frac{d y}{d x} - y^{- 2} \frac{d y}{d x} = 5$ $ye^(xy)+xe^(xy)dy/dx-y^-2dy/dx=5$

$\frac{d y}{d x} (x e^{x y} - y^{- 2}) = 5 - y e^{x y}$ $dy/dx(xe^(xy)-y^-2)=5-ye^(xy)$

$\frac{d y}{d x} = \frac{5 - y e^{x y}}{x e^{x y} - \frac{1}{y^{2}}}$ $dy/dx=(5-ye^(xy))/(xe^(xy)-1/y^2)$