anonymous
  • anonymous
Help:implicit differentiation with a fractions :(
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
anonymous
  • anonymous
multiplying everything by 36 you get; 2x^2+y^2=36 which maybe more clear for you, :)
anonymous
  • anonymous
hey! :) i tried this: x²/16 + y²/36 = 1 (1/16)x² + (1/36)y² = 1 2(1/16)x + 2(1/36)yy' = 0

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anonymous
  • anonymous
but so far im not sure what to do after because i know i get x/8 + ?? but then i can't really multiply (1/36) 2y*dy/dx ?
anonymous
  • anonymous
isn't the point here to find dy/dx?
anonymous
  • anonymous
let \(F(x,y)=2x^2+y^2=36\) then: \(dF(x,y)=4xdx+2ydy=0\) from here: \(y'=dy/dx=-4x/2y=-2x/y\)
anonymous
  • anonymous
huh? o.o
anonymous
  • anonymous
howd you get this?? let F(x,y)=2x2+y2=36
anonymous
  • anonymous
\(dF(x,y)\) is colled full differential it is: \(dF(x,y)=F_xdx+F_ydy\)
anonymous
  • anonymous
i never seen that before
anonymous
  • anonymous
\(F_x\) is partial derivative respect to x
anonymous
  • anonymous
ok
anonymous
  • anonymous
i know those, but where are the numbers coming from?
anonymous
  • anonymous
in what order, where did you take them?
anonymous
  • anonymous
i know they're given but how did you get 36 on the other side. and 2x^2+y^2let F(x,y)=2x2+y2=36
anonymous
  • anonymous
ok, other way: think that y is actualy a function of x, so it is y=y(x). Then: \(2x^2+y(x)^2=36\) now we differentiate keeping this in mind 4x+2yy'=0 now just solve for y'
anonymous
  • anonymous
2x2+y(x)2=36?
anonymous
  • anonymous
yes i know when to differentiate but how did you change the function to that
anonymous
  • anonymous
Ya you right I made a mistake. I was treating 36 like 32, :). Shold be. Rest of steps same. dw:1381788857534:dw|
anonymous
  • anonymous
|dw:1381789088155:dw|
anonymous
  • anonymous
i know I was going to tell you.
anonymous
  • anonymous
you can't have 36 at the end.
anonymous
  • anonymous
i saw a different way on yahoo, would be be able to help me with this one x²/16 + y²/36 = 1 (1/16)x² + (1/36)y² = 1 2(1/16)x + 2(1/36)yy' = 0 (x/8) + (y/18)y' = 0 (y/18)y' = -(x/8) y' = -(x/8)/(y/18) y' = -(x/8)(18/y) y' = -(9x/4y) y'(2) = -(9(2))/(4(5.2)) y'(2) = -18/20.8 y'(2) = -9/10.4 y'(2) = -0.865
anonymous
  • anonymous
I'm not really sure how they got the 4th step :/.... this part----> + (y/18)*dy/dx ??
anonymous
  • anonymous
2(1/16)x + 2(1/36)yy' = 0 (2/16)x+(2/36)yy'=0 (x/8) + (y/18)y' = 0
anonymous
  • anonymous
yes my question is exactly what you wrote up there .... |dw:1381789520628:dw|(x/8) + (y/18)y' = 0
anonymous
  • anonymous
(2/36)=1/18
anonymous
  • anonymous
how did it get simplified to y/18*dy/dx?
anonymous
  • anonymous
your answer is same, just simplify 2 in numerator with 36 in denominator
anonymous
  • anonymous
thanks! :) i dontkknow why i missed that silly stuff.
anonymous
  • anonymous
:)
anonymous
  • anonymous
then i plug in for x and y and find the common denominators rights?
anonymous
  • anonymous
yes
anonymous
  • anonymous
wait we're not done yet! :(
anonymous
  • anonymous
i got -.005379
anonymous
  • anonymous
the question is asking y'(1)
anonymous
  • anonymous
y' = -(9x/4y) so: substitute x=1, y=5.80948 y'(1)=-9/4(5.80948)
anonymous
  • anonymous
okay i got dy/dx=-.387298
anonymous
  • anonymous
got it thanks @myko :))!
anonymous
  • anonymous
yw

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