Help:implicit differentiation with a fractions :(

- anonymous

Help:implicit differentiation with a fractions :(

- chestercat

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- anonymous

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- anonymous

multiplying everything by 36 you get;
2x^2+y^2=36
which maybe more clear for you, :)

- 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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## More answers

- 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

isn't the point here to find dy/dx?

- 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

huh? o.o

- anonymous

howd you get this??
let F(x,y)=2x2+y2=36

- anonymous

\(dF(x,y)\) is colled full differential
it is: \(dF(x,y)=F_xdx+F_ydy\)

- anonymous

i never seen that before

- anonymous

\(F_x\) is partial derivative respect to x

- anonymous

ok

- anonymous

i know those, but where are the numbers coming from?

- anonymous

in what order, where did you take them?

- 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

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

2x2+y(x)2=36?

- anonymous

yes i know when to differentiate but how did you change the function to that

- anonymous

Ya you right I made a mistake. I was treating 36 like 32, :). Shold be. Rest of steps same.
dw:1381788857534:dw|

- anonymous

|dw:1381789088155:dw|

- anonymous

i know I was going to tell you.

- anonymous

you can't have 36 at the end.

- 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

I'm not really sure how they got the 4th step :/.... this part----> + (y/18)*dy/dx ??

- anonymous

2(1/16)x + 2(1/36)yy' = 0
(2/16)x+(2/36)yy'=0
(x/8) + (y/18)y' = 0

- anonymous

yes my question is exactly what you wrote up there .... |dw:1381789520628:dw|(x/8) + (y/18)y' = 0

- anonymous

(2/36)=1/18

- anonymous

how did it get simplified to y/18*dy/dx?

- anonymous

your answer is same, just simplify 2 in numerator with 36 in denominator

- anonymous

thanks! :) i dontkknow why i missed that silly stuff.

- anonymous

:)

- anonymous

then i plug in for x and y and find the common denominators rights?

- anonymous

yes

- anonymous

wait we're not done yet! :(

- anonymous

i got -.005379

- anonymous

the question is asking y'(1)

- anonymous

y' = -(9x/4y)
so: substitute x=1, y=5.80948
y'(1)=-9/4(5.80948)

- anonymous

okay i got dy/dx=-.387298

- anonymous

got it thanks @myko :))!

- anonymous

yw

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