anonymous
  • anonymous
Find dy/dx by implicit differentiation. 5+3x=sin(xy)^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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.
anonymous
  • anonymous
\[d/dx \sin (xy)^2 = d/dx sinu *d/dx (u) \] if we use the chain rule and (yx)^2=u
anonymous
  • anonymous
i got cos(xy)^2*2xy* 1*y*y' for that part but idk if its right

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anonymous
  • anonymous
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.
anonymous
  • anonymous
i no how to do the product rule, its just the implicit differientiation rule that confuses me
amistre64
  • amistre64
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
anonymous
  • anonymous
so when u differenciate the y wat does that become? y*y'?
amistre64
  • amistre64
yes, y becomes y' 6y^2 becomes 12y y'
anonymous
  • anonymous
oooooooooooooooooooo!!!!
amistre64
  • amistre64
xy becomes: x'y + xy' not that it helps here....maybe
amistre64
  • amistre64
...but it might lol
anonymous
  • anonymous
hmmm...I was thinking it would, since d(xy)/dx) = y+x*dy/dx?
amistre64
  • amistre64
2(sin(xy)) cos(xy) (x'y + xy') ??
anonymous
  • anonymous
so when i differenciate the whole thing it becomes 0+3=cos(xy)^2 * 2xy* (1)(y)(y')?
anonymous
  • anonymous
or not...
amistre64
  • amistre64
u^2 Du ; u = sin(t); t = xy
amistre64
  • amistre64
2sin(xy) cos(xy) (x'y + xy') does that help out?
anonymous
  • anonymous
yaa so thats for the right side only right
anonymous
  • anonymous
so its a chain rule within a chain rule?
amistre64
  • amistre64
yep; chain a chain lol
amistre64
  • amistre64
D(u) D(sin(t)) D(t) is what I get from it
anonymous
  • anonymous
ooooooo ok i get it!
anonymous
  • anonymous
so now i solve for y'
amistre64
  • amistre64
yep, x' = 1 so we can ignore those; and solve for y'
anonymous
  • anonymous
do i divide 2sinxy*cos(xy) on both sides?
amistre64
  • amistre64
its a good start :)
amistre64
  • amistre64
then subtract a "y" and divide it all again by "x"
anonymous
  • anonymous
so i got 3-y/2sin(xy)*cos (xy)*x
anonymous
  • anonymous
is that correct?
amistre64
  • amistre64
thats a little rough, not quite it....
anonymous
  • anonymous
awwww!!! lol
anonymous
  • anonymous
wat did i do wrong?
amistre64
  • amistre64
3 y ---------------- - -- x (2sin(xy)cos(xy)) x
anonymous
  • anonymous
where did the x under the 3 come from?
amistre64
  • amistre64
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'
anonymous
  • anonymous
oo i turn the x into a 1
amistre64
  • amistre64
x' = dx/dx = 1 like any good fracation does :)
anonymous
  • anonymous
ooooooooo ok i see wat i did wrong
amistre64
  • amistre64
you got 3-y/Bx
anonymous
  • anonymous
omg im gonna do exactly wat u did for the test.. change that whole part to a B so i wont make a mistake
amistre64
  • amistre64
lol ..... it is easier on the eyes fer sure :)
anonymous
  • anonymous
THANK U SOO MUCH!!!
anonymous
  • anonymous
UR A LIFE SAVER!!

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