Newcastle456
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Find dy/dx.

x= (2y) / (x^2 - y)

I appreciate there is an easier way to do this but I want to do it the hard way:

Differentiating with respect to x:

LHS= 1

RHS= Quotient Rule needs to be applied

Let 2y= u

(x^2-y)=v

du/dx= 2dy/dx

dv/dx= 2x - dy/dx

Quotient rule: v (du/dx) - u (dv/dx) / v^2

[(x^2 -y) (2 dy/dx) - 2y (2x - dy/dx)] / [(x^2-y) ^2]

[2x^2 dy/dx - 2y dy/dx - 4yx + 2y dy/dx] / [(x^2-y) ^2]

[2x^2 dy/dx - 4yx] / [(x^2-y) ^2]

Make LHS = RHS

1= [2x^2 dy/dx - 4yx] / [(x^2-y) ^2]


However, from here I can't get the simplied answer of dy/dx = (3x^2 - y) / (x+2)
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forgottensecret
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(Original post by Deep456)
Find dy/dx.


However, from here I can't get the simplied answer of dy/dx = (3x^2 - y) / (x+2)
http://www.wolframalpha.com/input/?i=differentiate+x%3D2y%2F%28x^2-y%29

I didn't get the simplified answer and from the looks of things neither did wolfram. I managed to get the second alternate form which wolfram provides:

\frac{dy}{dx}= \frac{x^4-2x^2y+y^2+4xy}{2x^2}

Edit: Yeah, just ignore all of what I've just wrote lol. Poster below got it
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Farhan.Hanif93
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(Original post by Deep456)
Find dy/dx.

x= (2y) / (x^2 - y)

I appreciate there is an easier way to do this but I want to do it the hard way:

Differentiating with respect to x:

LHS= 1

RHS= Quotient Rule needs to be applied

Let 2y= u

(x^2-y)=v

du/dx= 2dy/dx

dv/dx= 2x - dy/dx

Quotient rule: v (du/dx) - u (dv/dx) / v^2

[(x^2 -y) (2 dy/dx) - 2y (2x - dy/dx)] / [(x^2-y) ^2]

[2x^2 dy/dx - 2y dy/dx - 4yx + 2y dy/dx] / [(x^2-y) ^2]

[2x^2 dy/dx - 4yx] / [(x^2-y) ^2]

Make LHS = RHS

1= [2x^2 dy/dx - 4yx] / [(x^2-y) ^2]


However, from here I can't get the simplied answer of dy/dx = (3x^2 - y) / (x+2)
Note that:
x=\frac{2y}{x^2-y}

\implies y=\frac{x^3}{x+2}

\implies \frac{dy}{dx}=\frac{3(x+2)x^2 - x^3}{(x+2)^2}

Dividing numerator and denominator of the RHS by (x+2):

\implies \frac{dy}{dx}=\dfrac{3x^2-\frac{x^3}{x+2}}{x+2}

\implies \frac{dy}{dx} = \frac{3x^2-y}{x+2}

as required.
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