Question

????

Answer 1

So... derivative, ya?
How we start? Chain rule?\[\Large\rm e^{xy}(xy)'=0\]Understand that step? :d
what next?

Answer 2

So... derivative, ya?
How we start? Chain rule?\[\Large\rm e^{xy}(xy)'=0\]Understand that step? :d
what next?

Answer 3

Camila have we met before ?

Answer 4

No I don't think so. I would first subtract 1 from the xy and put it in front of the e correct?

Answer 5

Where are you from?

Answer 6

Recall your derivative for an exponential function,\[\Large\rm y=e^{u}\]\[\Large\rm y'=e^{u}u'\]Example:\[\Large\rm y=e^{2x}\qquad\to\qquad y'=e^{2x}(2x)'\]\[\Large\rm y'=e^{2x}(2)\]

Answer 7

So for our problem, I started out by taking the derivative of both sides,\[\Large\rm (e^{xy})'=(4)'\]\[\Large\rm e^{xy}(xy)'=0\]And from here, to finish up the differentiation procress, we need to apply our product rule.
See how we have x and y multiplying?
Ya product rule :o

Answer 8

How would you derive the e^xy though? Because the product rule is f'(x)g(x)+f(x)g'(x)

Answer 9

So no, we're not apply power rule.
We don't get an xy-1 or anything like that :)

We're applying exponential rule.
Exponential gives us the `same thing back` when we differentiate, just gotta remember to chain rule.

Like do you understand how to take this derivative? :o\[\Large\rm \left(e^{x^3}\right)'\]

Answer 10

I think it would be 3e^x^2?

Answer 11

No, we should get the `same thing back` when we differentiate the exponential.\[\Large\rm \left(e^{x^3}\right)'=e^{x^3}\text{__}\]See? Same thing as a result.
But we need to also chain rule: multiply by the derivative of the inner function.
The inner function being x^3 in this case.

Answer 12

\[\Large\rm \left(e^{x^3}\right)'=e^{x^3}(x^3)'\]\[\Large\rm \left(e^{x^3}\right)'=e^{x^3}(3x^2)\]

Answer 13

So back to our problem:\[\Large\rm \left(e^{xy}\right)'=e^{xy}\text{___}\]We get the same thing back...
but we're missing something, have to apply our chain rule.

Answer 14

Would you have to multiply it by its inverse?

Answer 15

You would multiply it by the `derivative of the inner function`.
The inner function is the stuff up in the exponent. The xy.

Answer 16

\[\Large\rm \left(e^{xy}\right)'=e^{xy}\text{___}\]\[\Large\rm \left(e^{xy}\right)'=e^{xy}(xy)'\]So now we'll `setup` our product rule:\[\Large\rm \left(e^{xy}\right)'=e^{xy}(x'y+xy')\]

Answer 17

\[\large\rm x'=1\]Derivative of x, with respect to x, is simply 1, ya?\[\Large\rm \left(e^{xy}\right)'=e^{xy}(x'y+xy')\]\[\Large\rm \left(e^{xy}\right)'=e^{xy}(1\cdot y+xy')\]

Answer 18

What do you think lady Cam? :d
too confusing?

Answer 19

I am very confused.

Answer 20

I know how to do the chain rule when it doesn't involve natural logs but now it got confusing.

Answer 21

We can actually use logs to simplify this problem BEFORE we take derivative.
Maybe that would help a little bit.

Answer 22

So like... if you take the natural log of each side at the start,\[\Large\rm \ln\left[e^{xy}\right]=\ln(4)\]Recall your log property that looks something like this,\[\rm \log(a^b)=b \log(a)\]The exponent can come out in front as a multiplier.
So for our problem, we apply that rule and get,\[\Large\rm (xy) \ln[e]=\ln(4)\]ln(e) will simplify to 1, (check that with your calculator if you need).
So we have:\[\Large\rm xy=\ln4\]This is the SAME PROBLEM.
It just might be easier to work with from here.

Answer 23

But, if working with logs is really confusing, we can maybe avoid taking that route.

Answer 24

It might be easier. So from xy=ln4 could I use product rule to get x+y=ln4 and then plug in my given values?

Answer 25

Woops, your derivative looks a little wonky.

Answer 26

First of all, let's deal with the right side.
You forgot to take the derivative of the right side.

(ln4)'=?

Answer 27

It equals 0

Answer 28

good good good, it's just a fancy looking constant. but it's still a constant.
so derivative is 0 :)

Answer 29

\[\Large\rm xy=\ln4\]\[\Large\rm (xy)'=(\ln4)'\]\[\Large\rm (xy)'=0\]Now the left side... hmmm
trying to figure out what you did there :d

Answer 30

I would recommend that you take the steps to first SETUP your product rule.
Don't just plug the stuff in until you're comfortable with it.\[\Large\rm (fg)'=f'g+fg'\]Yes?
So for our problem:\[\Large\rm (xy)'=x'y+xy'\]

Answer 31

x' is just the derivative of x, which is simply 1.\[\Large\rm (xy)'=y+xy'\]But we cannot treat y' in the same way.
This is our derivative term, dy/dx.

Answer 32

\[\Large\rm y+xy'=0\]So that finishes up the differentiation process at least, ya? :)

Answer 33

Now plug and chug

Answer 34

So its ln4+0=0 so it would just be ln4?

Answer 35

you're plugging in `1 for x`, and `ln4 for y`,
i'm not sure where you're getting that 0 from :o

Answer 36

Because you said xy'

Answer 37

you're trying to solve for y',
you don't plug anything into that.

Answer 38

\[\Large\rm \color{orangered}{y}+\color{orangered}{x}y'=0\]You're plugging your x and y values into these.

Answer 39

Ooooooo okay so it would be ln4+1=0 and then when you solve it would be -ln4+4?

Answer 40

\[\Large\rm \color{orangered}{y}+\color{orangered}{x}y'=0\]\[\Large\rm \color{orangered}{\ln4}+\color{orangered}{1}y'=0\]Your y' disappeared, I'm alil confused 0_o

Answer 41

Im so sorry I get really confused with math. Okay so you divided 0 by 1 and get 0 and subtract ln4 to get -ln4?

Answer 42

\[\Large\rm y'=-\ln4\]Yayyyy good job \c:/

Answer 43

\[\Large\rm \frac{dy}{dx}=-\ln 4\]

Answer 44

Thank you so much for helping me through this!!

Answer 45

But in your final steps there, don't be silly.
No reason to divide by 1 :)
Just ignore the 1.

1y' is the same as y'

Answer 46

Oh okay