How do you implicitly differentiate $11 = \frac{x}{1 - y e^{x}}$ $11=(x)/(1-ye^x)$ ?

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How do you implicitly differentiate $11 = \frac{x}{1 - y e^{x}}$ $11=(x)/(1-ye^x)$ ?

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Answer 1 · 2016-05-13T10:34:23

Consider $f (x, y (x)) = 0$ $f(x,y(x))=0$ taking the derivative gives
$\frac{d f}{d x} = (f_{x}) + (f_{y}) \frac{d y}{d x} = 0$ $(df)/dx = (f_x) +(f_y) (dy)/dx = 0$ so $\frac{d y}{d x} = - \frac{f_{x}}{f_{y}}$ $(dy)/dx=-f_x/f_y$ .
Now $f (x, y) = \frac{x}{1 - y e^{x}} - 11$ $f(x,y) = x/(1-ye^x)-11$
$f_{x} = \frac{d}{d x} (\frac{x}{1 - y e^{x}} - 11) = \frac{1 - e^{x} (x - 1) y}{{(e^{x} y - 1)}^{2}}$ $f_x = d/dx(x/(1-ye^x)-11) = (1-e^x(x-1)y)/(e^x y-1)^2$
$f_{y} = \frac{d}{d y} (\frac{x}{1 - y e^{x}} - 11) = \frac{y (x - 1) - e^{- x}}{{(e^{x} y - 1)}^{2}}$ $f_y = d/dy(x/(1-ye^x)-11) = (y(x-1)-e^-x)/(e^x y-1)^2$
finally $\frac{d y}{d x} = \frac{y (1 - x) - e^{- x}}{x}$ $(dy)/dx =(y(1-x)-e^-x)/x$

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How do you implicitly differentiate $11 = \frac{x}{1 - y e^{x}}$ $11=(x)/(1-ye^x)$ ?

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