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

##### 1 Answer

Aug 6, 2016

The slope is

#### Explanation:

This will be easier rewritten as:

#(e^y-e^x)^-1=C#

Now, we can use implicit differentiation (remember that each occurrence of

#-(e^y-e^x)^-2*d/dx(e^y-e^x)=0#

#-1/(e^y-e^x)^2(e^y*dy/dx-e^x)=0#

#(e^x-e^y(dy/dx))/(e^y-e^x)^2=0#

So, we want to solve for

#(e^-2-e^1(dy/dx))/(e^1-e^-2)^2=0#

Cross-multiplying:

#1/e^2-e(dy/dx)=0#

#dy/dx=1/e^3#