How do you implicitly differentiate $- y^{2} = e^{y} - \frac{y}{x}$ $-y^2=e^(y)-y/x$ ?

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How do you implicitly differentiate $- y^{2} = e^{y} - \frac{y}{x}$ $-y^2=e^(y)-y/x$ ?

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Answer 1 · 2016-04-18T13:02:55

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

Explanation:

As such functions cannot be explicitly written in the form $y = f (x)$ $y=f(x)$ , we differentiate them implicitly w.r.t $x$ $x$ or $y$ $y$ , but when we differentiate w.r.t. $y$ $y$ , we multiply by $\frac{d y}{d x}$ $dy/dx$ too (using chain rule).

Hence, differential of $- y^{2} = e^{y} - \frac{y}{x}$ $-y^2=e^y-y/x$ is

$- 2 y \frac{d y}{d x} = e^{y} \frac{d y}{d x} - \frac{x \cdot \frac{d y}{d x} - y \cdot 1}{x^{2}}$ $-2y dy/dx=e^y dy/dx-(x*dy/dx-y*1)/x^2$ or

$- 2 y \frac{d y}{d x} = e^{y} \frac{d y}{d x} - \frac{1}{x} \cdot \frac{d y}{d x} + \frac{y}{x^{2}}$ $-2y dy/dx=e^y dy/dx-1/x*dy/dx+y/x^2$ or

$\frac{1}{x} \cdot \frac{d y}{d x} - 2 y \frac{d y}{d x} - e^{y} \frac{d y}{d x} = \frac{y}{x^{2}}$ $1/x*dy/dx-2y dy/dx-e^y dy/dx=y/x^2$ or

$(\frac{1}{x} - 2 y - e^{y}) \frac{d y}{d x} = \frac{y}{x^{2}}$ $(1/x-2y-e^y)dy/dx=y/x^2$ or

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

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How do you implicitly differentiate $- y^{2} = e^{y} - \frac{y}{x}$ $-y^2=e^(y)-y/x$ ?

1 Answer

Explanation:

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How do you implicitly differentiate -y^2=e^(y)-y/x −y2=ey−yx-y^2=e^(y)-y/x ?

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Explanation:

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How do you implicitly differentiate $- y^{2} = e^{y} - \frac{y}{x}$ $-y^2=e^(y)-y/x$ ?