Question

Find dy/dx by implicit differentiation.
5+3x=sin(xy)^2

Answer 1

d/dx(5+3x) = d/dx(sin(xy)^2
so the left side is pretty obvious...I'm guessing its the RHS that's giving you trouble.

Answer 2

\[d/dx \sin (xy)^2 = d/dx sinu *d/dx (u) \] if we use the chain rule and (yx)^2=u

Answer 3

i got cos(xy)^2*2xy* 1*y*y' for that part but idk if its right

Answer 4

so d/dx (u) is where implicit differentiation comes in. You'll have to use two rules: the product rule of differentiation (d/dt (xy) = (x)'*y + x*(y)' ) and the implicit differentiation rule.

Answer 5

i no how to do the product rule, its just the implicit differientiation rule that confuses me

Answer 6

rules for implicit are exactly the same for explicit; only thing is you want to keep your x' and y' til the end then factor out your y'. your x'[dx/dx] will of course equal 1

Answer 7

so when u differenciate the y wat does that become? y*y'?

Answer 8

yes, y becomes y'

6y^2 becomes 12y y'

Answer 9

oooooooooooooooooooo!!!!

Answer 10

xy becomes:

x'y + xy' not that it helps here....maybe

Answer 11

...but it might lol

Answer 12

hmmm...I was thinking it would, since d(xy)/dx) = y+x*dy/dx?

Answer 13

2(sin(xy)) cos(xy) (x'y + xy') ??

Answer 14

so when i differenciate the whole thing it becomes 0+3=cos(xy)^2 * 2xy* (1)(y)(y')?

Answer 15

or not...

Answer 16

u^2 Du ; u = sin(t); t = xy

Answer 17

2sin(xy) cos(xy) (x'y + xy') does that help out?

Answer 18

yaa so thats for the right side only right

Answer 19

so its a chain rule within a chain rule?

Answer 20

yep; chain a chain lol

Answer 21

D(u) D(sin(t)) D(t) is what I get from it

Answer 22

ooooooo ok i get it!

Answer 23

so now i solve for y'

Answer 24

yep, x' = 1 so we can ignore those; and solve for y'

Answer 25

do i divide 2sinxy*cos(xy) on both sides?

Answer 26

its a good start :)

Answer 27

then subtract a "y" and divide it all again by "x"

Answer 28

so i got 3-y/2sin(xy)*cos (xy)*x

Answer 29

is that correct?

Answer 30

thats a little rough, not quite it....

Answer 31

awwww!!! lol

Answer 32

wat did i do wrong?

Answer 33

3 y
---------------- - --
x (2sin(xy)cos(xy)) x

Answer 34

where did the x under the 3 come from?

Answer 35

lets say B = (2sin(xy)cos(xy)) to clean this up...

3/B = y + xy'

(3/B) -y = xy'

(3/B)/x - y/x = y'

3/Bx - y/x = y'

Answer 36

oo i turn the x into a 1

Answer 37

x' = dx/dx = 1 like any good fracation does :)

Answer 38

ooooooooo ok i see wat i did wrong

Answer 39

you got 3-y/Bx

Answer 40

omg im gonna do exactly wat u did for the test.. change that whole part to a B so i wont make a mistake

Answer 41

lol ..... it is easier on the eyes fer sure :)

Answer 42

THANK U SOO MUCH!!!

Answer 43

UR A LIFE SAVER!!