Question

Implicit Differentiation
(x+y)^3 = x^3+y^3
find dy/dx at points (-1,1)

Answer 1

Im pretty confused. I'm starting out with 3(x+y)^2 * (1+dy/dx) on the left side. Currently redoing it

Answer 2

Yeah. I don't know where I'm going with this.

Answer 3

x=0
y=0

Answer 4

Thats not what the questions looking for

Answer 5

Thanks for trying :X

Answer 6

your initial work is fine

Answer 7

Currently on the left side I end up with 3(x+y)^2 * ( 1+ dy/dx) = 3x^2(dx/dx) + 3y^2(dy/dx) right side <-- am I going on the right path?

Answer 8

sorry I just looked at it and didn't think lol :D
I guess I need to pay more attention

Answer 9

consider this, since you are trying this with the d/dx notations:

d/dx [f] = df/dx

d/dx [y] = dy/dx

d/dx [x] = dx/dx, which is why its 1

Answer 10

youre doing great so far

Answer 11

I'm trying to do this step by step, so on the left side after simplifying it I ended up with
(3+3dy/dx) (x+y)^2.

Answer 12

The goal is to find dy/dx too by the way, at the point of (-1,1)

Answer 13

if we just let a ' define the derivative, the writeup goes

\[[(x+y)^3=x^3+y^3]'\]
\[[(x+y)^3]'=[x^3]'+[y^3]'\]
\[3(x+y)^2~[(x+y)]'=3x^2x'+3y^2y'\]
\[3(x+y)^2~(x'+y')=3x^2x'+3y^2y'\]
since x' = dx/dx = 1
\[3(x+y)^2~(1+y')=3x^2+3y^2y'\]
the rest is algebra to solve for y'

Answer 14

you can plug in the xy values at this stage if all the xy parts are confusing :) then algebra the y' out of it ... your call

Answer 15

So I plugged in (-1, and 1) for the last equation you gave me ( I got up to that too ) and I got derivative of y = -1. Is that correct? If It is Can you help me with a problem or two more?

Answer 16

notice that both sides have 3 as a common factor, so instead of trying to distribute it we can divide it off

\[\frac{\cancel3(x+y)^2~(1+y')=\cancel3x^2+\cancel3y^2y'}{\cancel3}\]
\[(x+y)^2~(1+y')=x^2+y^2y'\]
\[(-1+1)^2~(1+y')=1^2+(-1)^2y'\]
\[0=1+y'~:~y'=-1\]

Answer 17

Yep -1 :) Thank you! Can you help me out with one or two more? I had 10 problems and only these few confused/ made me double guess myself

Answer 18

i can spot you on 1 more ...

Answer 19

So for sqrtxy = x-2y at (4,1)

Answer 20

I got, 1/2(xy)^-1/2 * [x dy/dx + y dx/dx] ( for the left side )

Answer 21

good so far, an eyesore, but good

Answer 22

So then I simplified the first term, which I got 1/2sqrtxy then I multiplied it by the derivatives =
(x dy/dx + y) / 2sqrt xy

Answer 23

sqrt(xy) is more acceptable ascii notation; but go on

Answer 24

I then started working on the left side where I got 1-2 d(y)/d(x), Im unsure what to do now
Should I multiply both sides by 2sqrt(xy)?

Answer 25

xy' + y
------- = 1 - 2y'
2sqrt(xy)

if you want to do more work to it, thats up to you; but if all you needs a a value of y' at (4,1), id just plug and play it


4y' + 1
------- = 1 - 2y'
2sqrt(4)



4y' + 1
------- = 1 - 2y'
4


4y' + 1 = 4 - 8y'

now its more like a simple algebra problem .....

Answer 26

Alright, thanks, but it could eventually start working out if I did cancel out the denominator on the left side right?

Answer 27

of course, but i find that there comes a point when doing alot of "simplifying" to develop a formula for y' is just pointless is all

Answer 28

Haha thanks so much! Yeah I've been taught to just simplify to the end! Thank you for all your help!

Answer 29

youre welcome :) putting all that energy and time into simplifying when a text might only need a final number .... potato potato i spose lol

Answer 30

*a test....

Answer 31

Just a question, are you a real life math teacher or?