How do you use Implicit Differentiation of #ye^x=xe^y#?

Question

2616 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-08T08:27:35

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

First we take #d/dx# of each term.

#d/dx[ye^x]=d/dx[xe^y]#

#yd/dx[e^x]+e^xd/dx[y]=xd/dx[e^y]+e^yd/dx[x]#

#ye^x+e^xd/dx[y]=xd/dx[e^y]+e^y#

Using the chain rule, we know that:
#d/dx=d/dy*dy/dx#

#ye^x+dy/dxe^xd/dy[y]=dy/dxxd/dy[e^y]+e^y#

#ye^x+dy/dxe^x=dy/dxxe^y+e^y#

Now gather like terms together.
#dy/dxe^x-dy/dxxe^y=e^y-ye^x#

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

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