Question

PLEASE HELP
Determine the slope of the graph of x^2 = ln(xy) at the point (1, e).
1
2
e
e^3

Answer 1

Hey :)
So you'll need to differentiate implicitly,
which means we'll forgo looking for an explicit function y(x) before differentiating.
We'll just differentiate it as is.

It will involve some product rule and stuff like that.

Answer 2

\[\large\rm x^2=\ln(xy)\]Power rule on the left side,\[\large\rm 2x=\frac{d}{dx}\ln(xy)\]Do you remember your derivative for ln(x)?

Answer 3

is it 1/x?

Answer 4

Yes,\[\large\rm \frac{d}{dx}\ln(x)=\frac{1}{x}\]When we have more than just x inside of the log, we can generalize this rule with the `chain rule`:\[\large\rm \frac{d}{dx}\ln(stuff)=\frac{1}{stuff}(stuff)'\]

Answer 5

\[\large\rm 2x=\frac{d}{dx}\ln(xy)\]Let's try that with our problem.
Log(xy) becomes 1/(xy) when we differentiate it, ya?
And then chain rule tells us we have to multiply by the derivative of the inner function (xy),\[\large\rm 2x=\frac{1}{xy}\cdot(xy)'\]

Answer 6

Now we're taking the derivative of (xy).
We have a product, so we need to apply our product rule.\[\large\rm (xy)'=x'y+xy'\]

Answer 7

What do you think Sare? :O
Too confusing?

Answer 8

Yes, i think understand so far.

Answer 9

When I write x', I mean \(\rm \dfrac{d}{dx}x\)
So what is that derivative? :)

Answer 10

1?

Answer 11

Good.\[\large\rm (xy)'=(1)y+xy'\]\[\large\rm (xy)'=y+xy'\]our y' can not be simplied in a similar way.
y'=dy/dx, the derivative function.
This is what we're trying to solve for.

Answer 12

\[\large\rm 2x=\frac{1}{xy}\cdot(xy)'\]So this is what we have so far,\[\large\rm 2x=\frac{1}{xy}\cdot(y+xy')\]Oh actually they gave us a point to plug in,
I suppose we don't need to isolate our y' just yet.
Let's plug in the coordinate they gave us first.
It should make some of the math a bit easier on us.
\(\large\rm (x,y)=(1,e)\)

Answer 13

What do you get when you plug those x's and y's in?
Wanna give it a shot? :)

Answer 14

1/(1)(e) * (e+(1)(e)

Answer 15

Close! :)
We're not actually plugging anything for y'
\(\rm \dfrac{1}{1\cdot e}\cdot(e+1y')\)

And don't forget about the other side as well,\[\large\rm 2(1)=\dfrac{1}{1\cdot e}\cdot(e+1y')\]

Answer 16

We've evaluated y' at the point (1,e).
So this y' now represents the `slope` of the function at that point!
Solving for y' will answer the question for us.

Answer 17

wait i'm sorry why do you keep writing 2x. Doesn't the problem say x^2?

Answer 18

That username is so cool -_- lol

Answer 19

So what we did as a first step was: Take derivative of `both sides`.

Answer 20

ohh okay that why it's 2x.

Answer 21

\[\large\rm 2(1)=\dfrac{1}{1\cdot e}\cdot(e+1y')\]So your last step is to isolate the y'.
Think you can do that? :)

Answer 22

sure gimme one sec

Answer 23

y=1?

Answer 24

y' = 1 ?
Hmm that doesn't look quite right :o

Answer 25

Ignoring all of the 1's floating around,
we currently have this,\[\large\rm 2=\dfrac{1}{e}(e+y')\]So what's a good first step in simplifying?
Multiply both sides by e I suppose, ya?

Answer 26

Which leads to,\[\large\rm 2e=e+y'\]

Answer 27

now we divide right?

Answer 28

No silly >.<
We have e being `added` to y'.
Inverse of addition? :)

Answer 29

so subtract?

Answer 30

Yes.

Answer 31

e=2e-y or e=2-y

Answer 32

We would like to isolate the y',
we're trying to get everything on the left side of the equation.

So we don't actually want to mess with the 2e, it's in the correct place.
We'll undo the +e by subtracting e from each side.\[\large\rm 2e=e+y'\]\[\large\rm 2e-e=e-e+y'\]\[\large\rm 2e-e=y'\]Simply the left side.

Answer 33

so y=e

Answer 34

no wait 2?