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