What is the implicit derivative of #1= e^(xy) #?

2 Answers
Mar 6, 2016

#(dy)/dx=-y/x#

Explanation:

When we differentiate we have to use the chain rule in conjunction with the product rule.

The left had side is a constant #1# so its derivative with respect to #x# is #0#

For the right hand side we use the chain rule and the product rule.

#e^(xy)[y+x(dy)/dx]#

So together we have

#0=e^(xy)[y+x(dy)/dx]#

Distribute #e^(xy)#

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

Isolate term with #(dy)/dx#

#(dy)/dxxe^(xy)=-ye^(xy)#

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

#(dy)/dx=-y/x#

Mar 6, 2016

I like the question and the answer can be written #dy/dx = y/x#, but . . .

Explanation:

#1=e^(xy)# implies that #xy=0# which in turn implies that either #x=0# or #y=0#.

The graph of this equation is the pair of axes.

Here is the graph of #1=e^(xy)# using Socratic's graphing utility:

graph{1=e^(xy) [-10, 10, -5, 5]}