What is the slope of the tangent line of #e^(xy)-e^x/y = C #, where C is an arbitrary constant, at #(0,1)#?

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Answer 1 · 2017-09-19T22:54:28

The slope is #m=0#

If the curve goes through the point #(0,1)# then we have:

#e^(0xx1) -e^0/1 =C#

#1-1 = C#

#C = 0#

So the only curve of the family going through that point would be:

#e^(xy) = e^x/y#

Taking the logarithm of both sides we have:

#xy = x -lny#

Now if #y=1# this equation becomes #x=x#, which is satisfied by any value of #x#

If #y!=1# we have:

#x = lny/(1-y) !=0#

So the curve passing through the point #(0,1)# is the straight line #y=1#, and the tangent is the horizontal line itself with slope #m=0#.