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#