Question

Can somebody help me understand and solve this implicit differentiation problem? Find y^1 by implicit differentiation: (x^2)y=2x+siny

Answer 1

how far do you get?

Answer 2

and is that really a y^1?

Answer 3

find y' i think is what it means

Answer 4

oh no, its derivative, i just wasn't sure how to type it out

Answer 5

its a single quote :)

Answer 6

haha thanks

Answer 7

most of the time, implicits confusion is a result of not having a suitable understanding of a derivative to begin with. no offense

Answer 8

try this; what is the derivative of x^3 ?

Answer 9

yea, i admit i have problems with this, lol. i think, 3x^2?

Answer 10

close, but i never stated the derivative "with respect to x" so you instinctively threw out the derived bit

Answer 11

it is 3x^2 x'

Answer 12

try this one, derivative of: 2y^2

Answer 13

i'll try saying 4y?

Answer 14

youve got the thought down, but you keep throwing away your derived bit.

it is: 4y y'

Answer 15

oh shoot. yea, i guess i forget that part. how do u get y', again?

Answer 16

the confusion lies in this:
\[\frac{d}{dx}(x^3)=\frac{dx}{dx}\ 3x^2\]
but since we derive with respect to "x"; and dx/dx = 1. you are used to discarding it without a thought

Answer 17

\[\frac{d}{dx}(2y^2)=\frac{dy}{dx}\ 4y\]
the derived bit here doesnt go away since you have no idea what its value is

Answer 18

wow, u understand my train of thought better than i understand myself, lol

Answer 19

:) as long as you can retrain yourself to do derivatives correctly, implicits pose no problem

Answer 20

thats true. the thing that sucks about calculus is that it keeps building up on ur previous knowledge :(

Answer 21

\[x^2y=2x+sin(y)\]
\[\frac{d}{dx}[x^2y=2x+sin(y)]\]
\[\frac{d}{dx}[x^2y]=\frac{d}{dx}[2x]+\frac{d}{dx}[sin(y)]\]
\[\frac{d}{dx}[x^2y]=\frac{dx}{dx}2+\frac{dy}{dx}cos(y)\]
\[\frac{d}{dx}[x^2y]=2+\frac{dy}{dx}cos(y)\]
the only trick now is how to apply a product rule, but you already know the product rule right?

Answer 22

[fg]'=f'g+fg' right?

what happens when we change the letters?

[xy]'=x'y+xy'

Answer 23

nothing changes

Answer 24

\[\frac{d}{dx}[x^2y]=2+\frac{dy}{dx}cos(y)\]
\[\frac{d}{dx}[x^2]y+x^2\frac{d}{dx}[y]=2+\frac{dy}{dx}cos(y)\]
\[\frac{dx}{dx}2xy+x^2\frac{dy}{dx}1=2+\frac{dy}{dx}cos(y)\]
\[2xy+x^2\frac{dy}{dx}=2+\frac{dy}{dx}cos(y)\]
do you follow that?

Answer 25

the rest is just algebra to get to the dy/dx

Answer 26

oh so, would y'=(2-2xy)/x^2-cosy?

Answer 27

i believe so

Answer 28

yay! would u mind helping through an even more complex but simlar problem?

Answer 29

i got the time

Answer 30

thanks. here it is: find y" by implicit differentiation: x^4y^4=3

Answer 31

whoops, i meant to put x^4 + y^4 = 3

Answer 32

\[[x^4+y^4=3]'\]
\[[x^4]'+[y^4]'=[3]'\]
what would we do from here?

Answer 33

uh, taking derivative of each: 4x^3(x') + 4y^3(y')=1? I have a feeling its wrong...

Answer 34

its mostly right :) [3]'=0 since its just a constant

Answer 35

now solve it so y' is to one side

Answer 36

okay: y'=4x^3(x')/4y^3?

Answer 37

watch the algebra, simple mistakes will cost you on a test. and lets go ahead and consider x' to be 1, i just like to keep it there as a reminder of what happens while deriving.

y' = -4x^3/4y^3 = -x^3/y^3

Answer 38

oh i see

Answer 39

now if your goal is to have the 2nd derivative; then this is just a product rule

Answer 40

or a quotient rule; your pick :)

Answer 41

\[y'=-\frac{x^3}{y^3}\]or\[y'=-\left(\frac{x}{y}\right)^3\]or\[y'=-x^3y^{-3}\]

we have a few options to choose from and which ever one is easiest for you is the best

Answer 42

lets go with the first one :)

Answer 43

very well :)

\[[y'=-\frac{x^3}{y^3}]'\]
\[[y']'=-[\frac{x^3}{y^3}]'\]
\[y''=-\frac{y^3[x^3]'-[y^3]'x^3}{y^{3^2}}\]
\[y''=-\frac{y^3[x^3]'-[y^3]'x^3}{y^6}\]
how would we complete this?

Answer 44

sorry, i'm a little confused on the third step

Answer 45

the 3rd step is just the application of the quotient rule:
\[[\frac{t}{b}]'=\frac{bt'-b't}{b^2}\]

Answer 46

oh... gotcha

Answer 47

(y(3x^2)-(3y^2)x^3)/y^6?

Answer 48

needs some practice to it, but heres what I get:

\[y''=-\frac{y^3[x^3]'-[y^3]'x^3}{y^6}\]
\[y''=-\frac{y^3\ 3x^2\ x'-3y^2\ y'\ x^3}{y^6}\]
\[y''=-\frac{3x^2y^3-3x^3y^2\ y'}{y^6}\]
and since y' = -x^3/y^3 we can fil that in
\[y''=-\frac{3x^2y^3-3x^3y^2\ \frac{-x^3}{y^3}}{y^6}\]
\[y''=-\frac{3x^2y^3-3x^3\ \frac{-x^3}{y}}{y^6}\]
\[y''=-\frac{3x^2y^3+\frac{3x^6}{y}}{y^6}\]
\[y''=-\frac{\frac{3x^2y^4}{y}+\frac{3x^6}{y}}{y^6}\]
\[y''=-\frac{\frac{3x^2y^4+3x^6}{y}}{y^6}\]
\[y''=-\frac{3x^2y^4+3x^6}{y\ y^6}\]
\[y''=-\frac{3x^2y^4+3x^6}{y^7}\]
and hopefully i kept track of my signs

Answer 49

might simplify further to:
\[y''=-\frac{3x^2}{y^7}(-y^4-x^4)\]
or some such contrivial mess

Answer 50

that last simplification I did mess the signs on :/ lol

Answer 51

wow that was a really difficult problem