How do you differentiate #e^(xy)+y=x-1#?

2 Answers
Nov 23, 2017

Differentiate each term, put each term back into the equation, and then solve for #dy/dx#

Explanation:

Differentiate: #e^(xy)+y=x-1#

The first term:

Let #u = xy#, then #(du)/dx = (d(xy))/dx#

Use the product rule:

#(d(xy))/dx = (d(x))/dxy+x(d(y))/dx#

#(du)/dx = y + xdy/dx#

Use the chain rule:

#(d(e^(xy)))/dx = (d(e^u))/(du)(du)/dx#

#(d(e^(xy)))/dx = e^u(du)/dx#

#(d(e^(xy)))/dx = e^u(y + xdy/dx)#

#(d(e^(xy)))/dx = e^(xy)(y + xdy/dx)#

#(d(e^(xy)))/dx = ye^(xy) + xe^(xy)dy/dx#

The second term:

#(d(y))/dx = dy/dx#

The third term:

#(d(x))/dx = 1#

The fouth term:

#(d(-1))/dx = 0#

Put the terms back into the equation:

#ye^(xy) + xe^(xy)dy/dx + dy/dx = 1#

Solve for #dy/dx#:

#xe^(xy)dy/dx + dy/dx = 1- ye^(xy)#

#(xe^(xy) + 1)dy/dx = 1- ye^(xy)#

#dy/dx = (1- ye^(xy))/(xe^(xy) + 1)#

Nov 23, 2017

See explanation

Explanation:

First, we differentiate both sides with respect to x:

#d/dx e^(xy) + dy/dx = d/dx(x+1)#
#->d/dxe^(xy)+ y' = 1# (1)

For differentiating #e^(xy)#, recall that #d/dx e^u = (du)/dx *e^u#

With #u=xy, (du)/dx = d/dx (xy) #

Using the chain rule...

#d/dx (xy) = dx/dx*y + x*dy/dx = y + xy'#

Then back to (1), we get:
#d/dxe^(xy) + y' = 1 -> ye^(xy)+y'xe^(xy) + y' = 1#

We now move all terms without y' as a factor to the right hand side...

#->y'xe^(xy) + y' = 1 -ye^(xy)#

Factor out y' from the left hand side...

#-> y' (xe^(xy)+1)= 1 -ye^(xy)#

And divide both sides by #(xe^(xy)+1)..#

#y' = (1-ye^(xy))/(xe^(xy)+1#