Question

Find dy/dx (derivative) using Implicit Differentiation: xe^y=x-y

Answer 1

do you recall how to derive explicitly?

Answer 2

yes

Answer 3

then this is exactly the same, except for the fact that you are used to throwing away what I call, the derived bits

Answer 4

tell me, what is the derivative of x^3?

Answer 5

it's just that now we are solving for dy/dx and i get stumped when there is an "e" in the equation

Answer 6

thats fine, well get to that

Answer 7

3x^2

Answer 8

almost, i did not specify to derive with respect to x. and you instinctively threw out your derived bit

Answer 9

\[\frac{d}{d*}(x^3)=3x^2\ \frac{dx}{d*}\]
where * can be any "respect to"

Answer 10

this is the key to implicit derivatives

Answer 11

simply put; x^3 derives to: 3x^2 x'

Answer 12

what would you say y^3 derives to then?

Answer 13

3y^2 dy/dx

Answer 14

close; your assuming its with respect to "x"; and thats fine really, but not always accurate

Answer 15

y^3 derives to 3y^2 y'

Answer 16

its these derived bits that your used to tossing out since \(\frac{dx}{dx}=1\)

Answer 17

now lets try your posted problem

Answer 18

ok i see what you mean, go on

Answer 19

since this is with resepct to x, lets go about it like this

\[\frac{d}{dx}(xe^y=x-y)\]
\[\frac{d}{dx}(xe^y)=\frac{d}{dx}(x-y)\]
\[\frac{d}{dx}(xe^y)=\frac{d}{dx}(x)-\frac{d}{dx}(y)\]
\[\frac{d}{dx}(xe^y)=\frac{dx}{dx}-\frac{dy}{dx}\]
\[\frac{d}{dx}(xe^y)=1-\frac{dy}{dx}\]
\[\frac{d}{dx}(xe^y)=-\frac{dy}{dx}\]
we good so far?

Answer 20

the product rule is going to be applied to the left side there; and it works the same as it always does

Answer 21

\[\frac{d}{dx}(xe^y)=-\frac{dy}{dx}\]
\[\frac{d}{dx}(x)\ e^y+x\ \frac{d}{dx}(e^y)=-\frac{dy}{dx}\]
\[\frac{dx}{dx} e^y+x\ \frac{dy}{dx}e^y=-\frac{dy}{dx}\]
\[ e^y+x\ \frac{dy}{dx}e^y=-\frac{dy}{dx}\]
now we collect our dy/dx to one side so we can factor it out

Answer 22

\[e^y+x\ \frac{dy}{dx}e^y=-\frac{dy}{dx}\]
\[e^y=-\frac{dy}{dx}-x\ \frac{dy}{dx}e^y\]
\[e^y=\frac{dy}{dx}(-1-x\ e^y)\]
\[\frac{e^y}{(-1-x\ e^y)}=\frac{dy}{dx}\]
\[\frac{e^y}{-(1+x\ e^y)}=\frac{dy}{dx}\]
\[-\frac{e^y}{1+x\ e^y}=\frac{dy}{dx}\]

Answer 23

as you can see, after you get the deriving; the rest is just algebra

Answer 24

i agree with everything so far, but just one question...

where does the "1" go in the step where you had 1-dy/dx on the right side?

Answer 25

good eye, my brain said it was useless and i threw it out; which means that rest of the problem is wrong after that :)

same concept tho

Answer 26

haha okay thanks a lot though, i completely understand it now

Answer 27

good explanation!

Answer 28

\[e^y+x\ \frac{dy}{dx}e^y=1-\frac{dy}{dx}\]
\[e^y+x\ \frac{dy}{dx}e^y+\frac{dy}{dx}=1\]
\[e^y+\frac{dy}{dx}(x\ e^y+1)=1\]
\[\frac{dy}{dx}(x e^y+1)=1-e^y\]
\[\frac{dy}{dx}=\frac{1-e^y}{x e^y+1}\]
maybe lol