Question

3y+2x=sin(xy^2)
Find dy/dx
Can someone help me please

Answer 1

theres a product rule inside the sin(xy^2)

Answer 2

Implicit differentiation

Answer 3

yeah it comes out wrong i dont know why

Answer 4

left side i get 3yy'+2

Answer 5

and the right side i get cosy^2+cosy^2y'x

Answer 6

whats sham ?

Answer 7

i already know how to do this, it's calc 1 bat -.-

Answer 8

let's see you walk her through it. im busy atm

Answer 9

Alright

Answer 10

i get other questions similar to this right this one i keep getting it wrong

Answer 11

so the right side is 3yy'+2=cosy^2+cos2yy'x

Answer 12

than i combine like terms?

Answer 13

\[\frac{ d }{ dx }(2x+3y) = \frac{ d }{ dx }(\sin(xy^2))\]

Alright I'm going to show you this step by step, so you don't make a mistake next time you do a question like this, do you like the d/dx notation or do you want me to use y'?

Answer 14

use y' and thanx :)

Answer 15

\[(2x+3y)'=(\sin(xy^2))'\]

I just saw you mention y' so I deleted everything I did lol, I'll restart though this way, it's fine. Haha.

Answer 16

lol sorry

Answer 17

\[2+3y'=\cos(xy^2)*(xy^2)'\] chain rule

\[2+3y' = \cos(xy^2) *((x)'(y^2)+(y^2)'(x))\]

\[2+3y' = \cos(xy^2)*(y^2+2yy'x)\]

\[2+3y' = \cos(xy^2)y^2+2xcos(xy^2)yy'\]
Expand


\[2+3y'-2xcos(xy^2)yy'=\cos(xy^2)y^2\]
Subtracting \[2xyy'\cos(xy^2)\] from sides

\[3y'-2xcos(xy^2)yy'=-2+\cos(xy^2)y^2\]
Subtracting 2.

\[y'=\frac{ -2+\cos(xy^2)y^2 }{ 3-2xcos(xy^2)y }\]
Dividing both sides by \[3-2xycos(xy^2)\]

Answer 18

Hope that helps :p

Answer 19

Thanx!

Answer 20

Your welcome :)