Question

7=xy-e^(xy) dy/dx=?

Answer 1

Simplify as much as possible

Answer 2

\(\large\color{black}{ 7 =xy~-~e^{xy} }\)

Answer 3

Do you want to do logarithmic differentiation or just do the chain for e^xy, and product for xy ?

Answer 4

I would suggest my second approach.

Answer 5

by the way, I tried to solve this and I got to \[\frac{ dy }{ dx }=\frac{ -y+e^{xy} }{ x-e^{xy}}\]

Answer 6

I used the second approach

Answer 7

let's see.

Answer 8

Sorry the top is -y+ye^(xy)

Answer 9

\(\Large\color{black}{7 =xy~-~e^{xy} }\)


\(\Large\color{black}{~ 0 =1\times y+x\times \frac{dy}{dx}~-~e^{xy}\times(1\times y+x\times \frac{dy}{dx})}\)


\(\Large\color{black}{~ 0 =y+x\frac{dy}{dx}~-~e^{xy}\times (y+x\frac{dy}{dx})}\)



\(\Large\color{black}{~ 0 =y+x\frac{dy}{dx}~-~ye^{xy}+e^{xy}x\frac{dy}{dx}}\)

Answer 10

\(\Large\color{black}{~ -y+ye^{xy} =x\frac{dy}{dx}~+e^{xy}x\frac{dy}{dx}}\)


\(\Large\color{black}{~ -y+ye^{xy} =(x+e^{xy}x)\frac{dy}{dx}}\)

Answer 11

One mistake it should be -e^xy x dy/dx

Answer 12

I need to find out where the x came from. Just a second