Question

Calculus: For 4x^(3)-3xy^(2)+y^(3)=28 find dy/dx at the point (3,4) on the curve. Can someone go step by step with me on how to do this please?

Answer 1

yes easy.. first find the dy/dx of that equation.. bring all terms on one side..

you would get an expression in x and y, then subsititute 3,4 for x and y to get the final value!

Answer 2

can you differentiate this implicitly, or are you having trouble with that?

Answer 3

yeahh i think im having trouble with that; and okay but before i find the dy/dx, do i have to set it equal to zero or to y?

Answer 4

when you take the derivative the right side will become zero, because the derivative of any constant is zero

Answer 5

you cannot put x and y terms on separate sides, you need to directly differentiate.. !!

Answer 6

quickly answer these what are the derivatives of
4x^3
3xy^2
y^3 ?? wrt x

Answer 7

ohh okay! so then after finding the derivatives, do i plug in both the x and y values they gave me?

Answer 8

12x^2

Answer 9

3y^2

Answer 10

and do i have to use the product rule for the second?

Answer 11

no, that is the point, you need to use the chain rule here

Answer 12

yes, and the product rule for the second

Answer 13

so try again, what is the derivative of
3xy^2
with respect to x

Answer 14

im sorry i actually haven't memorized the product rule yet :(

Answer 15

well you sort of need to to do these.
d/dx(f(x)g(x))=f(x)g'(x)+f'(x)g(x)

Answer 16

http://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/implicitdiffdirectory/ImplicitDiff.html Here are some good examples of implicite differentiation.

Answer 17

okay before i do, since there is a 3, isn't that a constant so i don't have to include that?

Answer 18

thank you eashmore!

Answer 19

the 3 can stay outside if you want because for any constant C we have
d/dx[Cf(x)]=Cf'(x)
constants are funny that way

Answer 20

is it 2xy+y^2?

Answer 21

for what? xy^2
almost, but you didn't differentiate y implicitly. You need the chain rule:
d/dx[f(g(x))]=f'(g(x))g'(x)

Answer 22

what exactly does implicitly mean? and so do i use the chain rule for just the y^2 or the whole xy^3?

Answer 23

whole

Answer 24

derivative of y wrt x is dy/dx.. so if you have to find derivative of y^2 wrto x it is 2ydy/dx!

Answer 25

when it is the derivative w/ respect to x, we need to differentiate y implicitly.
Look what Mashy said. d/dx(y^2)=2yy', not just 2y

Answer 26

that is the use of the chain rule. You need to use it everywhere there is a y (or whatever variable you're not differentiating w/respect to)

Answer 27

by the way the answer is 1/10

Answer 28

ah okay so let me try this again

Answer 29

and 1/10 isn't an answer choice :P oh calculus...

Answer 30

i got 10/3

Answer 31

ok i will check it again

Answer 32

is that a choice?

Answer 33

the answer choices are 5/2, 4/3, 5/12, -7/6, and 2/3

Answer 34

oh sorry I get 5/2
my bad... arithmetic :P

Answer 35

lol np thanks! but i know for a fact ill see this one my exam and unfortunately i won't have OpenStudy with me >.<

Answer 36

No, so you better try to see what you're missing so we can work on it.
try the derivative of
3xy^2 again. This time keep the 3. Don't forget the product and chain rule.

Answer 37

ah okay i got 6xy+3xy^2 but i can't figure out where to use the chain rule in this

Answer 38

close again, but not quite...
that x on the right should be gone, we took the derivative of it.
plus we need the chain rule: when we take the derivative of y we need to multiply what we get by dy/dx
wanna try again?

Answer 39

differentiaion of 3xy^2 is read as .. 3 ( x into differentiation of y^2 + y^2 into differentiation of x).. now tell the answer :P

Answer 40

here it is slow:

d/dx(3xy^2)=3[d/dx(xy^2)]=3(y^2dx/dx+2xy*dy/dx)
=3y^2+6xyy'

where y' is the derivative dy/dx
try to follow that step by step

Answer 41

I can't see my LaTeX font right now so if somebody else wants to type it more clearly...

Answer 42

what's the *?

Answer 43

times

Answer 44

d/dx(3xy^2)=3[d/dx(xy^2)]=3[y^2(dx/dx)+2xy(dy/dx)]
=3y^2+6xyy'

Answer 45

\[{d \over dx} 3xy^2 = 3 \left[ { d \over dx}(xy^2) \right] = 3 \left[ y^2 \left(dx \over dx \right) + 2xy \left( dy \over dx \right) \right] = 3y^2 + 6xyy''\]

Answer 46

thanks^

Answer 47

AH!!! NOW I GOT IT!

Answer 48

cool, so what do you have for the derivative wrt x of
4x^(3)-3xy^(2)+y^(3)=28
???
don't forget to take the derivative of everything, and use the chain rule
...and product rule.

Answer 49

eeekk i don't think this is right at all. -12x^(2)/ [-3Y^(2)+6xy+3y^(2)]

Answer 50

first of all, what happened to the other side of our equation?

Answer 51

well i copied the way my calc teacher told us to do it (which he is the worst). dy/dx= -12x^(2)/ [-3Y^(2)+6xy+3y^(2)]

Answer 52

???
that is madness, you should be doing it the way you see above as done by eashmore
what is d/dx of...
1)4x^3
2)-3xy^2
3)y^3
take each individually

Another thing, you forgot y' from the chain rule. That is critical to our problem. Whenever we are taking the derivative wrt x of y we need to wind up with a y'.

Answer 53

i know... im one of the worst in the class... anyways, 1) 12x^2 2) 3y^2+6xyy' 3) 3y^2?

Answer 54

good except for the last one... what is it missing?

Answer 55

y'?

Answer 56

right, because we took the derivative of a y term.

Answer 57

oh sorry, actually you forgot that we made the second one negative as well
1)4x^3 1)12x^2
2)-3xy^2 2)-6xyy'-3y^2
3)y^3 3)3y^2y'
so now that we have our derivatives we can add up those terms.
what will be on the right side of our equation.
Write the whole equation if you can.

Answer 58

was i suppose to get this?
12x^2-3y^2-6xyy'+3y^2y'=0 ?

Answer 59

yes, nice!
now do you think you can solve it for y' ?

Answer 60

ah! that's what i wrote on my paper!(except i had a plus 6 instead of minus) so after that is where i messed up, solving for y'

Answer 61

perhaps, if you say so. try it again!

Answer 62

okay so so far i have
\[-6xyy'+3y ^{2}y'=-12x ^{2}+3y ^{2}\]

is that a good start?

Answer 63

good so far
now factor out y' from the left...

Answer 64

or note that you could divide everything by 3 first if you want to cut out a few numbers

Answer 65

so then as the final thing

\[y'= -4x ^{2}+y ^{2}\div -2xy+y ^{2}\]

Answer 66

the division was suppose to be something over something but i didn't know how to do that

Answer 67

I can't see that font right now anyway, but it looks like
y'=(-4x^2+y^2)/(-2xy+y^2)
right?

Answer 68

ohh okay but yeah that's what it looks like

Answer 69

So now we just plug in our point:
(x,y)=(3,4)

Answer 70

ah so that's how we will get 5/2?

Answer 71

hopefully :)

Answer 72

ah yay yay yay! :D thank you ever so much!!! <3

Answer 73

no problem
things to take away:
1)practice the product rule
2)practice the chain rule
not that long of a list.
Well congratulations, any questions?

Answer 74

for this question no im find :) and yes im definitely going to have to practice those

Answer 75

fine*

Answer 76

well, see ya later, good luck :)

Answer 77

see ya! thanks again! (: