Question

Find dy/dx .... xy^2 = cos x

Answer 1

implicit differentiation?!

Answer 2

implicit differentiation
do you know product and chain rule ?

Answer 3

Is that like d/dx (xy^2) = d/dx cos x

Answer 4

Product rule: \[\huge f'(x)g(x) +f(x)g'(x)\]

Answer 5

\[(x \frac{ d }{ dx } (y ^{2}) + y ^{2} \frac{ d }{ dx } (x)\]

Answer 6

Chain rule, here to make it more clear:

\[\huge h(x) = f[g(x)]\]

\[\huge h’(x) = f’[g(x)]g’(x)\]

Answer 7

but because there is a y what do you do?

Answer 8

dy/dx

Answer 9

don't you have to solve for dy/dx because in the end dy/dx will always be on the left hand side of the equation..

Answer 10

Yes, that's the point :p

Answer 11

I would just solve out right but nahhhhhhhhh where is the learning ? :P

Answer 12

If it's easier, just use y' notation.

Answer 13

NO! that's BAD S:

Answer 14

who wants me to just end the misery type 1

Answer 15

wait...im still confused

Answer 16

hint: every time you take the derivative of y you have to write dy/dx next to it

Answer 17

What do you get when you use product rule on xy?

Answer 18

type 1 to end misery ^^.

Answer 19

Yeah pretty much what usuki said, dy/dx wherever you're finding derivative of y.

Answer 20

another hint: you need the chain rule and product rule... if you haven't mastered it yet go practice then come back to this

Answer 21

for example

y^2

derivative of y

2y dy/dx

Answer 22

The rules are also written above ^^

Answer 23

\[xy ^{2} = \cos x\]
\[\frac{ d }{ dx } (xy ^{2}) = \frac{ d }{ dx } (\cos x)\]
\[(x \frac{ d }{ dx } (y ^{2}) + y ^{2} \frac{ d }{ dx } (x) = -\sin x\]