Question

find dy/dx by implicit differentiation for
x^1/2 + y^1/2 = 16
I kind of understand the guidelines for it, but at the same time I'm ehh about it.

Answer 1

ok so since they want you to use implicit differentiation, you don't need to solve for "y" first...

Answer 2

basically you just do this:
1/2x^-1/2 + 1/2y^-1/2dy/dx = 0

Answer 3

i worked it out and i checked the answer and it says that it should be \[-\sqrt{y/x} \] im not sure how that happened.

Answer 4

you have to solve for y prime by making the dy/dx stand for y prime.

Answer 5

all you do is take the derivatives of each piece like you normally would, but when you take the derivative of "y", you need to remember the dy/dx part...because you're eventually solving for dy/dx

Answer 6

do you understand what i did so far?

Answer 7

the derivative of x^1/2 is 1/2x^-1/2

Answer 8

yeah that i understand

Answer 9

the 1/2 in the front is cancelled out? that just dissapeared. thats what i wanted to know.

Answer 10

Im saying that for a later step, umm im going through the work for the answer online and i didnt understand what happened there.

Answer 11

\[D[x^{1/2} + y^{1/2} = 16]\]
\[D[x^{1/2}] +D[y^{1/2}] = D[16]\]
\[\frac{1}{2x^{1/2}}x' +\frac{1}{2y^{1/2}}y' = 0\]
\[\frac{1}{2y^{1/2}}y' = -\frac{1}{2x^{1/2}}x'\]
\[y' = -\frac{2y^{1/2}}{2x^{1/2}}x'\]
\[y' = -\frac{y^{1/2}}{x^{1/2}}x';\ \text{and x' = 1 with respect to x}\]
\[y' = -\frac{y^{1/2}}{x^{1/2}}\]

Answer 12

my internet sucks right now :(

Answer 13

it's ok im still trying to understand amistre, i feel like i understand this , like as far as the guidelines that my textbook explain. Now even better with amistre's explanation and everyone elses. But if I continue doing this, i'm still going to make some sort of mistake... -.-

Answer 14

implicits are nothin special, or any different from explicits

Answer 15

tell me, what is the derivative of: 3x^2 ?

Answer 16

6x

Answer 17

close, but no.

6x x'

i never said the derivative with respect to x. and what you are used to doing is throwing this precious little derived bit away since it equals 1; with respect to itself

Answer 18

what is the derivative of: 3y^2 ? ; or 3r^2, or 3m^2?

Answer 19

the rules dont change and never have; you are just used to throwing out the derived bit by habit instead of sorting it out in the end

Answer 20

so it would be 6y y' ?

Answer 21

yes

Answer 22

hmm maybe because i just stated doing this and my teacher didnt explain that, but she never told me to do that and my text book doesnt show that. (atleast so far... )

Answer 23

it wont show that either ....

Answer 24

oh so thats just something i should know? hmmm

Answer 25

its something you should be taught :) and once taught, all the rest of the enigma tends to go away

Answer 26

the product rule is still the product rule right?

[xy]' = x'y+xy'

Answer 27

yes lol

Answer 28

if this is with respect yo y; y'=1; if with respect to x, x'=1; if respect to time; neither equals 1

Answer 29

these derived bit are the workings of the chain rule; always poping out little derived bits

Answer 30

keep them and sort thru them in the end and you will never go wrong

Answer 31

So I would use the chain rule for lets say x^2y+y^2x =-2 I've been kind of stuck on how to approach this one

Answer 32

the chain rule is at work; but not the full blown method that youre used to it being here

Answer 33

i see product, and power rules

Answer 34

D[x^2 y + y^2 x = -2]

D[x^2 y] + D[y^2 x] = D[-2]

--------------------------------------

D[x^2 y] = x'^2 y+x^2 y' = 2x x' y + x^2 y'

D[y^2 x] = y^2' x+y^2 x' = 2y y' x+ y^2 x'

D[-2] = 0
----------------------------


2x x' y + x^2 y' + 2y y' x+ y^2 x' = 0

and since x'=1 with respect to x we can toss them

2xy + x^2 y' + 2y y' x+ y^2 = 0

and solve for y'

Answer 35

x^2 y' + 2xy y' = -2xy - y^2

y' (x^2 + 2xy) = -2xy - y^2

y' = (-2xy - y^2)
----------
(x^2 + 2xy)

Answer 36

or neatly:

\[D[x^2 y + y^2 x = -2]\]
\[D[x^2 y] + D[y^2 x] = D[-2]\]
--------------------------------------

\[D[x^2 y] = x'^2 y+x^2 y' = 2x x' y + x^2 y'\]
\[ D[y^2 x] = y'^2 x+ y^2 x' = 2y y' x+ y^2 x' \]
\[D[-2] = 0\]
----------------------------

\[2x x' y + x^2 y' + 2y y' x+ y^2 x' = 0\]
and sort

Answer 37

Ohh boy, and to think i was once good at math, this has made me think twice about it. So for some reason the whole x' y' is totally throwing me off.

Answer 38

they are simply place holders in a sense

Answer 39

you can reformat them as \(y'=\cfrac{dy}{d*}\) and \(x'=\cfrac{dx}{d*}\)

and then fill in the * part afterwards really

Answer 40

if we say we want this with respect to x then we get: \(y'=\cfrac{dy}{dx}\) and \(x'=\cfrac{dx}{dx}=1\)

Answer 41

the best thing to do is to pick a method that works best for you and own it :)

Answer 42

maybe thats whats throwing me off, i use dy/dx and dx/dx which is 1 instead of y' and it just looks weird to me. because based on the work u showed me i think i figured how to do it. but im not sure

Answer 43

woah not because i meant but * lol