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

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Answer 1 · 2017-07-20T22:22:22

#1#

Making #z = x-y# we have

#ze^z=C# or #z= W(C)# where #W(cdot)# is the Lambert Function.

then #z = z_0 = W(C)#

Now #x_0 = x-y# and #dx=dy# or #dy/dx=1#

and finally

#dy/dx=1#

Answer 2 · 2017-07-20T22:48:08

The slope of the tangent line is #m=1#

Differentiate the equation implicitly:

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

#(x-y)d/dx(e^(x-y)) + (1-y')e^(x-y) = 0#

#(x-y)(1-y')e^(x-y) + (1-y')e^(x-y) = 0#

#(x-y+1)(1-y')e^(x-y) = 0#

As #e^t != 0# this implies:

#(x-y+1)(1-y') = 0#

We have then two possible solutions:

#(1)" " (1 -y') = 0#

#dy/dx = 1#

#y = x + a#

or:

#(2)" "(x-y+1) = 0#

#y = x+1#

which is just the particular case of #(1)# for #a = 1#.

Thus:

#y=x+a#

is the general solution, which is always a straight line of slope #m=1# for any #C# coincident with ts tangent.