Question

implicit derivative of sin(2xy)

Answer 1

I found the answer to be (-ycos(2xy))/(xcos(2xy))

Answer 2

Your answer would have to be in terms of d/dy=

Answer 3

Yes? That's the whole point of implicit differentiation.

Answer 4

or y'

Answer 5

Chain rule and product rule.

Answer 6

No not y', y' would just be a regular derivative. This is implicit differentiation

Answer 7

Yes I know, I'm telling you this problem requires chain rule and product rule though. Just with y'.

Answer 8

Yes

Answer 9

First thing you have to do is multiply the equation by d/dx

Answer 10

@jim_thompson5910

Answer 11

I believe u= 2xy Then you'd have

cosu(dericative of u) Which you'd have to do implicit differentiation ad the product rule

Answer 12

In the original problem, is sin(2xy) equal to anything?

Answer 13

^ that's also why I was confused with this

Answer 14

just f(x)

Answer 15

so perhaps they meant to say

y = sin(2xy)

Answer 16

derive both sides to get

y = sin(2xy)

dy/dx = cos(2xy) * ( d/dx[ 2xy ] )

dy/dx = cos(2xy) * ( 2y + 2x*dy/dx )

and then isolate dy/dx

Answer 17

Isolating dy/dx gives you...

dy/dx = cos(2xy) * ( 2y + 2x*dy/dx )

dy/dx = 2y*cos(2xy) + cos(2xy)*2x*dy/dx

dy/dx - cos(2xy)*2x*dy/dx = 2y*cos(2xy)

dy/dx [1 - 2x*cos(2xy) ] = 2y*cos(2xy)

dy/dx = ( 2y*cos(2xy) )/(1 - 2x*cos(2xy) )

so here's what it looks like in a cleaner format

\[\Large \frac{dy}{dx} = \frac{2y\cos(2xy)}{1 - 2x\cos(2xy)}\]

Answer 18

Optionally, you can think of

dy/dx = cos(2xy) * ( 2y + 2x*dy/dx )

as

z = q * ( 2y + 2x*z )

where z = dy/dx and q = cos(2xy)

Then solve for z