C4 Implicit Differentiation - HelpWatch

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Thread starter 8 years ago
#1
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)
0
8 years ago
#2
(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: Edit: Yeah, just ignore all of what I've just wrote lol. Poster below got it 0
8 years ago
#3
(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:   Dividing numerator and denominator of the RHS by :  as required.
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