Question

find dy/dx using implicit differentiation: xe^y-ye^x=5

Answer 1

The implicit differentiation formula is,
\[F(x,y)=xe^y-ye^x-5\Rightarrow \frac{dy}{dx}=-\frac{\partial F/\partial x}{\partial F/\partial y}\]
Could you continue from this point?

Answer 2

I'm still pretty confused, could you go furthur?

Answer 3

further*

Answer 4

I don't think he is referring to partial derivatives here, @John_ES

Answer 5

Use product rule for xe\(^y\) and also with ye\(^x\)

Answer 6

It is merely derivatives and tedious algebra solving for \(\sf \color{purple}{\frac{dy}{dx}}\)

Answer 7

Oh, well. It's true, there's another way to do the problem. I don't ask if @dn80919 has seen something about partial derivatives. Do you see them? If not, better follow the method @abb0t is explaining ;).

Answer 8

hmmm..how would you take derivative of xe^y?

Answer 9

Product rule: \(\sf \color{blue}{\frac{d}{dx}f(x)g(x)}\) = \(\sf \color{red}{ f(x)\frac{d}{dx}[g(x)]+g(x)\frac{d}{dx}[f(x)]}\)

Answer 10

You should get: \(\sf \color{orange}{\large xe^y~\frac{dy}{dx}+e^y}\)

Answer 11

Are you sure of that? I think he should apply differentiation and not derivation. I mean,
\[dx\cdot e^y+xe^ydy-dy\cdot e^x-ye^xdx=0\] And extracto common factor,
\[dx(e^y-ye^x)+dy(xe^y-e^x)=0\]So,
\[\frac{dy}{dx}=-\frac{e^y-ye^x}{xe^y-e^x}\]

Answer 12

differentiation is the process of deriving the derivative
to say "derive" means nothing unless you say what it is that you are deriving.

Answer 13

@zzr0ck3r congrats on becoming green :)

Answer 14

ty sir:)