Question

When you implicitly take the derivative of a function that has a term with both x AND y in it, how do you know if it is dx/dx or dy/dx? Ex. (x^2)y

Answer 1

If that makes sense at all

Answer 2

dx/dy or dy/dx is make more sense than dx/dx or dy/dy, right?

Answer 3

What?

Answer 4

Read your question again!!

Answer 5

I meant dy/dx, sorry

Answer 6

Not dy/dy

Answer 7

yes, sanity is back :)

Answer 8

Lol

Answer 9

The physical problem I am working on is x^2y+xy^2=6

Answer 10

And they want me to find dy/dx

Answer 11

Usually, when they ask you take implicitly, they ask for dy/dx. In special case, they will specifically name out.

Answer 12

ok, just take derivative as usual.

Answer 13

Thats the issue, idk which terms are dx/dx and which are dy/dx

Answer 14

What is the change rule?

Answer 15

You mean the product rule???

Answer 16

We use dy/dx notation, so i don't really understand what all of the primes mean

Answer 17

prime is derivative. That's why you confused.

Answer 18

I am REALLY confused

Answer 19

We are supposed to implicitly take the derivative

Answer 20

So there should be dy/dx and dx/dx in there

Answer 21

Ok

Answer 22

So instead of actually taking the derivative of anything, you just use prime?

Answer 23

Of course, why do we confuse ourselves by dx/dy, dx/dx, dy/dy, dy/dx.....???

Answer 24

you see, even wolframalpha (a machine) can understand that (x^2)' is \(\dfrac{d}{dx}x^2\)

Answer 25

http://www.wolframalpha.com/input/?i=%28x^2%29%27%3D

Answer 26

No, (x^2)' would be 2x(dx/dx)

Answer 27

Which is 2x(1)=2x

Answer 28

dx/dx =1,

Answer 29

Ok, thank you!!!