Question

Find dy/dt by implicit differentiation and evaluate the derivative at the point (1,1)
x^3 + y^3= 2xy

Answer 1

I kind of started it but I got lost really quickly

Answer 2

is the question is finding dy/dx or dy/dt ?? Please check it again

Answer 3

oh Im sorry! dy/dx

Answer 4

can you do
d/dx x^3 ?

Answer 5

3x^2?

Answer 6

I got 3x^2+ 3y^2 * dy/d

Answer 7

yes
now
d/dx y^3

it is almost the same
3y^2 dy/dx (the dy/dx must go in there)

Answer 8

dy/dx* I just dont know what the 3xy turns into

Answer 9

for x*y use the product rule

Answer 10

d(xy) = x * d/dx y + y * d/dx x

Answer 11

what is the first and what is the second function? is he first 2x and the second y?

Answer 12

the order will not matter

Answer 13

In case of dy/dx, you can solve by implicit differentiation as following

\[d(x^3 + y^3) = d(2xy)\]
\[3x^2 + 3y^2.y \prime = 2 (y + xy \prime)\]
\[y \prime (3y^2-2x) = 2y - 3x^2 \rightarrow y \prime = \frac{ 2y - 3x^2 }{ 3y^2 - 2x }\]

Evaluate at (1,1)
\[y \prime = \frac{ 2 - 3 }{ 3 - 2 } = -1\]

Answer 14

and I would leave the 2 out front (it will multiply the answer)
2 * d(xy)

Answer 15

So is the derivative of 2xy 2x * y' + 2y?
If it isnt, I have no clue then

Answer 16

that looks good

Answer 17

So then does it become dy/dx=-3x^2-3y^2 - 2x

Answer 18

in detail:
\[ 2 \frac{d}{dx} (xy) = 2\left(x\frac{d}{dx} y + y \frac{d}{dx}x\right) \\
2 x \frac{dy}{dx} + 2y \frac{dx}{dx} \\
2 x \frac{dy}{dx} + 2y
\]

Answer 19

Idk how to clean it up and simplify it

Answer 20

you sub in (1,1)

Answer 21

you should (at the start) have
\[ 3 x^2 + 3y^2\frac{dy}{dx} = 2x \frac{dy}{dx} + 2y \]
I would plug in x=1, y=1 and simplify to find dy/dx

Answer 22

like how do you find out what dy/dx is because that is the first part of the problem? I have dy/dx on both sides of the equation

Answer 23

isnt is supposed to be dy/dx= something?

Answer 24

btw
***
So then does it become dy/dx=-3x^2-3y^2 - 2x
***
I don't know how you got this.

**
isnt is supposed to be dy/dx= something?
***

Yes.
you treat dy/dx just like an unknown variable (e.g. "x") that you "solve for"

But it might be faster to first put in 1 for x and y first...

Answer 25

okay so then I get 6 dy/dx= 2dy/dt+ 2?

Answer 26

Do you understand how we got to
\[ 3 x^2 + 3y^2\frac{dy}{dx} = 2x \frac{dy}{dx} + 2y \]
?
now sub in (1,1) (which means x=1, y=1)
\[ 3 + 3 \frac{dy}{dx}= 2 \frac{dy}{dx} + 2\]
if it helps, think of this as
\[ 3 + 3x = 2x + 2 \]
solve for x, and rename x as \( \frac{dy}{dx} \)

Answer 27

***okay so then I get 6 dy/dx= 2dy/dt+ 2?***

your algebra is not correct.

Answer 28

how?

Answer 29

3+3x=2x+2

can you solve this for x ?

Answer 30

add -3 to both sides, then add -2x to both sides
then simplify

Answer 31

x=-1

Answer 32

now all I did was temporarily rename \(\frac{dy}{dx}\) as x to make the equation look simpler.

in other words, you would do the same thing with
\[ 3+3\frac{dy}{dx}=2\frac{dy}{dx}+2 \]

Answer 33

add -3 to both sides, add -2 \(\frac{dy}{dx}\) to both sides

Answer 34

So when you find the derivative at (1,1) the answer is dy/dx= -1?

Answer 35

yes, the answer is
\[ \frac{dy}{dx}= -1 \]at the point (1,1)

Answer 36

So there isnt two different answers for this problem?

Answer 37

the differentiation and the derivative at the point is all one answer?

Answer 38

If you scroll up and look at linh's post, he shows how to solve for dy/dx (he used y' for dy/dx) in terms of x and y, then subbed in x=1 and y=1 at the end.

Answer 39

**
the differentiation and the derivative at the point is all one answer?
***

the general answer for dy/dx is in terms of x and y
see linh's post.
once you replace x and y with 1, you get dy/dx at that point... it will be a number. (-1 in this case)

Answer 40

oh okay thank you

Answer 41

can you do the algebra to solve for dy/dx in terms of x and y, starting with
\[3 x^2 + 3y^2\frac{dy}{dx} = 2x \frac{dy}{dx} + 2y \]

it looks a bit more complicated... But reading the question, I think they want to first solve for dy/dx, and then substitute in (1,1)
(I skipped that step, and substituted first, because I was solving for dy/dx at (1,1) )