Question

Find dy/dx sqrt(x+y)=cuberoot(x-y) i got stuck please help Find dy/dx sqrt(x+y)=cuberoot(x-y) i got stuck please help @Mathematics

Answer 1

i got to this point
(1+dy/dx)(3(x-y)^2/3) = (1-dy/dx)(2(x+y)^1/2)

Answer 2

what is it you need to do?

Answer 3

its implicid differentiation just finding dy/dx to simplest form

Answer 4

oh, cause that could be confused for a diffy q if read the wrong way

Answer 5

i dont know how to separate dy/dx from what i got assuming its right.. algrebra =(

Answer 6

sorry about that i would type it better if i know how to use the equation functions

Answer 7

(x+y)^(1/2) = (x-y)^(1/3)

1/2 (x'+y') (x+y)^(-1/2) = (1/3) (x'-y') (x-y)^(-2/3); x'=1

1/2 y' (x+y)^(-1/2) = -(1/3) y' (x-y)^(-2/3)

1/2 y' (x+y)^(-1/2) + (1/3) y' (x-y)^(-2/3) = 0

y' [1/2 (x+y)^(-1/2) + (1/3) (x-y)^(-2/3)] = 0

well, when we divide off the baggage we get zero, so let me dbl chk

Answer 8

i see what i did, the ascii made me lose track lol

Answer 9

\[(x+y)^{1/2} = (x-y)^{1/3}\]
\[\frac{[x+y]'}{2(x+y)^{1/2}} = \frac{[x-y]'}{3(x-y)^{2/3}}\]
\[\frac{y'}{2(x+y)^{1/2}} = \frac{-y'}{3(x-y)^{2/3}}\]
hmmm

Answer 10

\[\frac{1+y'}{2(x+y)^{1/2}} = \frac{1-y'}{3(x-y)^{2/3}}\]
thats better lol

Answer 11

yup!

Answer 12

i threw out my x's alltogether at first ;)

Answer 13

hahah i saw that

Answer 14

this is where i dont know how to get y' on one side

Answer 15

\[\frac{1+y'}{2(x+y)^{1/2}} = \frac{1-y'}{3(x-y)^{2/3}}\]
\[(1+y')(3(x-y)^{2/3}) = (1-y')(2(x+y)^{1/2})\]
\[\frac{1+y'}{1-y'} = \frac{2(x+y)^{1/2}}{3(x-y)^{2/3}}\]
well, its to one side :)

Answer 16

\[\sqrt{x+y}=\sqrt[3]{x-y}\]\[(1/2)(x+y)^{-1/2}(1+y')=(1/3)(x-y)^{-2/3}(1-y')\]\[(3/2)(x+y)^{-1/2}+(3/2)(x+y)^{-1/2}y'=(x-y)^{-2/3}-(x-y)^{-2/3}y'\]\[(3/2)(x+y)^{-1/2}-(x-y)^{-2/3}=[-(3/2)(x+y)^{-1/2}-(x-y)^{-2/3}]y'\]\[y'={(3/2)(x+y)^{-1/2}-(x-y)^{-2/3}\over-(3/2)(x+y)^{-1/2}-(x-y)^{-2/3}}\]I don't know and I don;t want to try to simplify this,it's quite ugly, but I think I did ok, yeah? somebody check my work please.

Answer 17

looks about right

http://www.wolframalpha.com/input/?i=implicit+derivative+%28x%2By%29%5E%281%2F2%29+%3D+%28x-y%29%5E%281%2F3%29

Answer 18

id be curious to see if this is the same result :)
\[\sqrt{x+y}=\sqrt[3]{x-y}\]
\[(\sqrt{x+y})^3=x-y\]
\[(\sqrt{x+y})^2\sqrt{x+y}=x-y\]
\[(x+y)\sqrt{x+y}=x-y\]
\[\sqrt{x+y}=\frac{x-y}{x+y}\]
\[x+y=\frac{x^2-2xy+y^2}{x^2+2xy+y^2}\]
\[x^3+3x^2y+3xy^2+y^3=x^2-2xy+y^2\]
\[[x^3+3x^2y+3xy^2+y^3=x^2-2xy+y^2]'\]
\[3x^2+6xy+3x^2y'+3y^2+6xyy'+3y^2=2x-2y-2xy'+2y\]
\[3x^2y'+6xyy'+2xy'=-3x^2+2x-6xy-6y^2\]
\[y'(3x^2+6xy+2x)=-3x^2+2x-6xy-6y^2\]
\[y'=\frac{-3x^2+2x-6xy-6y^2}{3x^2+6xy+2x}\]

Answer 19

should i do it turing test way>

Answer 20

I think my answer is correct, but totally unsimplified. If in doubt you can always cheat and do it the wolfram way.

http://www.wolframalpha.com/input/?i=implicit+derivative+%28x%2By%29%5E%281%2F2%29+%3D+%28x-y%29%5E%281%2F3%29

Answer 21

so ill just have to simplify your answer and get similar to the wolfram?