Question

Use imclicit differentiation to find y'
xe^y - 10x + 3y = 0

I'm having trouble with understand how to do this... :/

Answer 1

Take your best guess and I'll help you out. I love this stuff!

Answer 2

-.-"
*brain gets fried*
erm... is derivative xe^y -> e^y ???

Answer 3

The example I like to show is:

\[y=x^2\]
algebra doesn't change how one variable changes with respect to the other. \[y^{1/2}=x\] take the derivative of both sides with respect to x\[\frac{1}{2}y^{-1/2}*\frac{dy}{dx}=1x^0*\frac{dx}{dx}\]
So all I did there was the chain and power rules. Notice how that simplifies to:
\[\frac{dy}{dx}=2y^{1/2}=2x\] which is what we knew all along. If any of that is confusing, ask!

Answer 4

...?
how does that apply to my problem? :/

Answer 5

I am showing you how implicit differentiation works.

Answer 6

*nods*
but idk how derivative of xe^y works

Answer 7

You just need to use the product rule.

Answer 8

???
xy'e^y + e^y ???

Answer 9

\[\frac{d}{dx}(xe^y)=\frac{d}{dx}(x)e^y+x \frac{d}{dx}(e^y)\]

Answer 10

Yep you got it. =)

Answer 11

y' = (10 - e^y) / (xe^y + 3) ???

Answer 12

Is that your final answer?

Answer 13

Wait, yay you're right jiggly ^.^

Answer 14

Gj!

Answer 15

YAYYYY THANK YOU!!! :D

Answer 16

This stuff is awesome, as kai mentioned haha.

Answer 17

.-. your smartie panties

Answer 18

tee-hee, panties .-.

Answer 19

I hate the d/dx notation though haha, I'm more of y' kind of guy :)

Answer 20

yrsh easier to type :P

Answer 21

I guess I just think it's more "true" to what's going on. Whatever works lol.