find dy/dx ln(x2y) = (e^xy)/y

[kamurphy]

1. Find dy/dx ln(x2y) = (e^xy)/y
   
   

 

[Deleted member 4993]

2. Find dy/dx ln(x2y) = (e^xy)/y

Your post is ambiguous! Please use grouping symbols, e.g. [{()}], to clarify order of operations - and repost including your work.

 

[kamurphy]

find dy/dx

[ln(x^2y)]=[(e^(xy))/y]

 

[Deleted member 4993]

find dy/dx

[ln(x^2y)]=[(e^(xy))/y]

y*[ln(y)] + y*[2 * ln(x)] = e^(xy)

Start differentiating - using chain rule and product rule.

 

[kamurphy]

answer?

I got the answer as [(-2+(e^(xy)x)y^2] / [x(-e^(xy)-y+e^(xy)xy]
not sure if this is right or not?

 

[stapel]

I got the answer as [(-2+(e^(xy)x)y^2] / [x(-e^(xy)-y+e^(xy)xy]
not sure if this is right or not?

I'm getting something very different. Please reply showing your steps. Thank you!