Question

Use implicit differentiation to find dy/dx where x^2y = 1 + e^y.

Answer 1

i can get as far as... 2xy * y(dy/dx)

then i don't know how to do the e^y

Answer 2

use chain rule,
d/dx(e^y) = e^y d/dx(y)

Answer 3

which is e^y y'
got this ?

Answer 4

hmmm nope. i don't see it...

Answer 5

d/dx of e^x is just e^x
but here y is the function of x
so u need chain rule

Answer 6

\[\frac{ dy }{ dx } = \frac{ -2xy }{ x^2}\]

that's as far as i can get...

Answer 7

that whole e to the y throws me off so much...

Answer 8

no, u need to take care of e^y also
and \(x^2y\) needs a product rule

Answer 9

kira that's the wrong questoin...

Answer 10

i know i do. i'm just saying i don't know how to take care of e^y

Answer 11

and that i said
e^y needs chain rule because its function of x
d/dx (e^y) = e^y d/dx(y) = e^y dy/dx
did u get this ?
if not which step ?

Answer 12

no that just confuses me... can you give another example?

Answer 13

In that case,
d/dx (e^y) = d/dy (e^y) * dy/dx = e^y * dy/dx

Answer 14

yeah that's what hartnn said...

Answer 15

sure
d/dx (sin y) = cos y d/dx(y) = cos y dy/dx

Answer 16

If z = ln y and we want to find dz/dx
dz/dx = 1/y * dy/dx

Answer 17

dz/dx = dz/dy * dy/dx

Answer 18

u got two examples :)

Answer 19

well i haven't used cos or sin yet so that's a little confusing... let me see...

Answer 20

d/dx( y^3) = 3y^2 d/dx (y) = 3y^2 dy/dx

Answer 21

want more examples ?

Answer 22

sure i'll take one more.

i do get it with number exponents. like 2y^2 would be 4y(dy/dx) right?

Answer 23

yes, absolutely correct :)

Answer 24

then what is up with e^y?! :(

Answer 25

i like your avatar by the way

Answer 26

first , whats d/dx of e^x

Answer 27

thanks :)

Answer 28

e^x

Answer 29

right , but now here its not e^x , its e^y
where y is the function of x
so using chain rule, we have to differentiate y after we differentiate e^y
so we get like this
d/dx (e^y) = e^y . d/dx (y)

Answer 30

...okay...

Answer 31

so after diff. u get
2xy + x^2 dy/dx = e^y dy/dx
got this ?

Answer 32

now isolate dy/dx

Answer 33

i know i'll have 2xy over x^2 on the right siide. and dy/dx on the left, but i don't know what to do with e^y(dy/dx) that's on the right...

Answer 34

*negative 2xy

Answer 35

2xy + x^2 dy/dx = e^y dy/dx

x^2 dy/dx - e^y dy/dx = -2xy
(x^2 - e^y ) dy/dx = -2xy
can u finish it off ?

Answer 36

\[\frac{ -2xy }{ x^2 - e^y }\]

Answer 37

what confused me was the double dy/dx... i didn't think of it like factoring.

Answer 38

thats correct :)

Answer 39

damn you math!! but thank you hartnn!!

Answer 40

welcome ^_^