Question

e^(xy) = 2x + 5y + 21. Find dy/dx

Answer 1

Understand how we will deal with the left side?\[\Large \left(e^{(stuff)}\right)' \quad=\quad e^{(stuff)}(stuff)'\]

Answer 2

\[\Large \left(e^{xy}\right)' \quad=\quad e^{xy}(xy)'\]

Answer 3

Hmm looks like we need the product rule, yes?

Answer 4

Yeah, I get that, I tried and I got \[dy/dx (e^(xy)) (y+x(dy/dx))=2+5(dy/dx)\] Which as far as I can tell can't be simplified?

Answer 5

We're going to want to group the (y')s.
Hmm why is there a dy/dx in front of the exponential term?

Answer 6

\[\Large e^{xy}(y+xy')\quad=\quad2+5y'\]Your work looks good besides that little part.

Answer 7

To put multiple things inside of an exponent, the code requires you to use curly braces.
e^{xy}

just something to keep in mind ^^

Answer 8

I may not understand this, but isn't the derivative of e^xy \[(dy/dx)(e^(xy)) \] and then the rest?

Answer 9

Oh okay, my bad, first time using this site.

Answer 10

No, we only throw on a dy/dx when we take the derivative of y directly.

Answer 11

So why doesn't the derivative of e^xy have a (dy/dx) ? There's a y in that part of it?

Answer 12

Oooh I see.

Answer 13

It will have a dy/dx eventually :) after chain rule finishes hehe

Answer 14

inside the brackets, like you did correctly.

Answer 15

Thanks a ton, that means I can simplify then. I wasn't understandingwhere to put (dy/dx). Okay, that clears up a lot. Thanks.

Answer 16

cool :)

Answer 17

Alright. Thanks again. How do I close the question?

Answer 18

Umm I think there is a button near the top of your thread to close it.
It's in blue if I recall.

Answer 19

Oh okay I see. Anyway to honor you more than just with the best response button?

Answer 20

No not really :)
You can go to someone's profile and leave a fan testimonial if you feel they've been really helpful. Not a big deal though.
I'm not here for brownie points hehe.

Answer 21

oh okay then. ^.^ I appreciate the help. You have a good one.