How can I find the derivatives of the implicitly set function y=y(x), set with the equation #xe^y + ye^x - e^(xy) = 0#? the answer I have is y' = ye^xy -e^y -ye^x/xe^y + e^x -xe^xy

2 Answers
Mar 27, 2018

#dy/dx = (ye^(xy)-ye^x-e^y)/(xe^y + e^x -xe^(xy))#

Explanation:

Given: #xe^y + ye^x - e^(xy) = 0#

Differentiate each term with respect to x:

#(d(xe^y))/dx + (d(ye^x))/dx -(d(e^(xy)))/dx = (d(0))/dx" [1]"#

The first term requires the use of the product rule:

#(d(uv))/dx = (du)/dxv+u(dv)/dx#

Where #u = x and v = e^y#, then #(du)/dx =1 and (dv)/dx = (d(e^y))/dx#

We need the chain rule for #(d(e^y))/dx#:

#(d(e^y))/dx = (d(e^y))/dydy/dx#

#(d(e^y))/dx = e^ydy/dx#

Substituting into the product rule:

#(d(xe^y))/dx = e^y+xe^ydy/dx" [2]"#

Substitute equation [2] into equation [1]:

#e^y+xe^ydy/dx + (d(ye^x))/dx -(d(e^(xy)))/dx = (d(0))/dx" [1.1]"#

The next term, also, requires the use of the product rule:

#(d(uv))/dx = (du)/dxv+u(dv)/dx#

Where #u = y, and v = e^x#, then #(du)/dx = dy/dx and #(dv)/dx = e^x#

Substituting into the product rule:

#(d(ye^x))/dx = dy/dxe^x+ye^x" [3]"#

Substitute equation [3] into equation [1.1]:

#e^y+xe^ydy/dx + dy/dxe^x+ye^x -(d(e^(xy)))/dx = (d(0))/dx" [1.2]"#

If we let #u = xy#, then we can use the chain rule for the next term:

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

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

But we shall need the product rule to compute #(du)/dx#

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

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

Reverse the substitution for u:

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

Use the distributive property:

#(d(e^(xy)))/dx = ye^(xy) +xe^(xy)dy/dx" [4]"#

Substitute equation [4] into equation [1.2] (remember to distribute the leading -1):

#e^y+xe^ydy/dx + dy/dxe^x+ye^x -ye^(xy) -xe^(xy)dy/dx = (d(0))/dx#

The derivative of 0 is 0:

#e^y+xe^ydy/dx + dy/dxe^x+ye^x -ye^(xy) -xe^(xy)dy/dx = 0#

Move all of the terms that do NOT contain #dy/dx# to the right:

#xe^ydy/dx + dy/dxe^x -xe^(xy)dy/dx = ye^(xy)-ye^x-e^y#

Factor out #dy/dx# on the left:

#(xe^y + e^x -xe^(xy))dy/dx = ye^(xy)-ye^x-e^y#

Divide both sides by #(xe^y + e^x -xe^(xy))#:

#dy/dx = (ye^(xy)-ye^x-e^y)/(xe^y + e^x -xe^(xy))#

Mar 27, 2018

#y'=-1/x#

Explanation:

notice that #ye^x# and #e^xy# are same.Therefore #ye^x-e^xy=0#

this will give you #xe^y=0#

applying product rule #(f.g)'=f.g'+f'.g#

so #y'xe^y+e^y=0#
rearrange the equation,
#y'xe^y=-e^y#
#y'=-e^y/(xe^y)#
y'=#-1/x#

alternatively,
#ln(xe^y)=0#
#lnx +ln e^y=0#
#lnx+y=0#
#y=-ln x#
#y'=-1/x#