What is the slope of the tangent line of $\frac{1}{e^{y} - e^{x}} = C$ $1/(e^y-e^x) = C$ , where C is an arbitrary constant, at $(- 2, 1)$ $(-2,1)$ ?

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What is the slope of the tangent line of $\frac{1}{e^{y} - e^{x}} = C$ $1/(e^y-e^x) = C$ , where C is an arbitrary constant, at $(- 2, 1)$ $(-2,1)$ ?

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Answer 1 · 2016-08-06T16:55:22

The slope is $\frac{1}{e^{3}}$ $1/e^3$ .

Explanation:

This will be easier rewritten as:

${(e^{y} - e^{x})}^{- 1} = C$ $(e^y-e^x)^-1=C$

Now, we can use implicit differentiation (remember that each occurrence of $y$ $y$ will put the chain rule in effect):

$- {(e^{y} - e^{x})}^{- 2} \cdot \frac{d}{d x} (e^{y} - e^{x}) = 0$ $-(e^y-e^x)^-2*d/dx(e^y-e^x)=0$

$- \frac{1}{{(e^{y} - e^{x})}^{2}} (e^{y} \cdot \frac{d y}{d x} - e^{x}) = 0$ $-1/(e^y-e^x)^2(e^y*dy/dx-e^x)=0$

$\frac{e^{x} - e^{y} (\frac{d y}{d x})}{{(e^{y} - e^{x})}^{2}} = 0$ $(e^x-e^y(dy/dx))/(e^y-e^x)^2=0$

So, we want to solve for $\frac{d y}{d x}$ $dy/dx$ when $(x, y) = (- 2, 1)$ $(x,y)=(-2,1)$ .

$\frac{e^{- 2} - e^{1} (\frac{d y}{d x})}{{(e^{1} - e^{- 2})}^{2}} = 0$ $(e^-2-e^1(dy/dx))/(e^1-e^-2)^2=0$

Cross-multiplying:

$\frac{1}{e^{2}} - e (\frac{d y}{d x}) = 0$ $1/e^2-e(dy/dx)=0$

$\frac{d y}{d x} = \frac{1}{e^{3}}$ $dy/dx=1/e^3$

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What is the slope of the tangent line of $\frac{1}{e^{y} - e^{x}} = C$ $1/(e^y-e^x) = C$ , where C is an arbitrary constant, at $(- 2, 1)$ $(-2,1)$ ?

1 Answer

Explanation:

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What is the slope of the tangent line of 1/(e^y-e^x) = C 1ey−ex=C1/(e^y-e^x) = C , where C is an arbitrary constant, at (-2,1)(−2,1)(-2,1)?

1 Answer

Explanation:

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What is the slope of the tangent line of $\frac{1}{e^{y} - e^{x}} = C$ $1/(e^y-e^x) = C$ , where C is an arbitrary constant, at $(- 2, 1)$ $(-2,1)$ ?