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

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Answer 1 · 2016-08-06T16:55:22

The slope is #1/e^3#.

This will be easier rewritten as:

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

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

#-(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 #dy/dx# when #(x,y)=(-2,1)#.

#(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#