Question

use partial derivatives to find dy/dx if y=f(x) is determined implicitly by the given equation.

2x^3+x^2y+y^3=1

Answer 1

\[2x^3+x^2y+3=1\]

Answer 2

you dont use partial for implicits

Answer 3

thats what the directions on the hw say lol

Answer 4

implicits imply that y changes as a result of x

partials are when y changes independant of x

Answer 5

i don't know what you want me to say lol, this is directly from the book..

Answer 6

then get a new book :) there is no way I see that this can be answered as is with any degree of intelligence ...

Answer 7

as is, it would simply be an implicit differentation

Answer 8

the answer is :
\[- (6x^2+2xy) / (x^2+3y^2)\]
if that sheds any light?
cuz i'm completely lost.. this chapter is on the chain rule?

Answer 9

6x^2+2xy+x^2y'+3y^2 y' = 0

Answer 10

yep, its just an implicit ....

Answer 11

move the non y's to the right; then factor out your y' and divide off the slag

Answer 12

you've lost me ^.^

Answer 13

do you know how to do implicit differentation?

Answer 14

no :(

Answer 15

do you know how to do regular differentation?

Answer 16

yes!

Answer 17

then you know implicit, youve just not been taught that you know it yet

Answer 18

your used to getting rid of the x' bit

Answer 19

for example; tell me, what is the derivative of 2x?

Answer 20

2

Answer 21

nope, it 2x'

i never said "with respect to x"

Answer 22

what is the derivative of 5y^2 ?

Answer 23

5y^2' ? lol

Answer 24

10y y' ...

Answer 25

whats the derivative of sin(2t) ?

Answer 26

now you've lost me lol
2sin(2t)'?!

Answer 27

you learn from the start how to derive things with respect to x ...
\[\frac{d}{dx}4x^3\to \frac{dx}{dx}12x^2\]
since dx/dx = 1 you think its not there

Answer 28

2cos(2t) t'

Answer 29

assume that we dont give a "with respect to ....

\[\frac{d}{d*}4x^3\to\ \frac{dx}{d*}12x^2\]

know we dont know what x' is equal to do we?

Answer 30

\[\frac{d}{d*}4xy^2\to\ \frac{d4}{d*}xy^2+\frac{dx}{d*}4y^2+\frac{dy}{d*}4x2y\]
\[x'\ 4y^2+y'\ 8xy\]

Answer 31

if *=x , its simplifies to

\[4y^2+y'\ 8xy\]

Answer 32

all implicits are doing is not throwing out the derived bits

Answer 33

product rule is still product rule

power rule is still power rule

and all the other rules youve learned are still all the other rules; nothing changes.

Answer 34

so 2x^3 is 6x^2x' ?

Answer 35

yes

Answer 36

so if its asking for dy/dx, what is it asking me to derive? o.O

Answer 37

yes; but derive it with the bits left in; then since this defines /dx all the x' s = 1

Answer 38

all the y' s are just dy/dx

Answer 39

\[x^2y\]
derive it with power and product rule

Answer 40

mostly product rule; but power when needed :)

Answer 41

uhh......
x^2x'+2x'y'?!? o.O

Answer 42

almost, keep your wits about you

first split it into the product rule

x^2' y + x^2 y' ; then derive the parts

2xx' y + x^2 y' ; and since this is /dx then x' = 1

2xy + x^2 y'

Answer 43

derive this one now

y^3

Answer 44

3y^2y'?

Answer 45

exactly

Answer 46

and the constant after the equals sign derives to 0 like every good constant needs to

Answer 47

put it all together and slove it out for y'

Answer 48

2xy + x^2 y'
why isn't there an x' in this?

Answer 49

there was, but read the next line; since dx/dx = 1 we can simplify it

Answer 50

oohhhh okay that makes sense! i think i get it now! thank you!!!!

Answer 51

practice keeping all your derived bit in ; and then the variables that match the /d* part go to one :)

good luck