Find $\frac{d y}{d x}$ $dy/dx$ if $y^{2} = x {(x - 1)}^{2}$ $y^2=x(x-1)^2$ ?

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Find $\frac{d y}{d x}$ $dy/dx$ if $y^{2} = x {(x - 1)}^{2}$ $y^2=x(x-1)^2$ ?

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Answer 1 · 2017-05-25T00:53:05

$\frac{d y}{d x} = \frac{(x - 1) (3 x - 1)}{2 y}$ $dy/dx = ((x-1)(3x-1))/(2y)$

Explanation:

We have:

$y^{2} = x {(x - 1)}^{2}$ $y^2 = x(x-1)^2$

Differentiating LHS implicitly, and RHS using the product rule along with the chain rule we get:

$(\frac{d}{d y} y^{2}) (\frac{d y}{d x}) = (x) (\frac{d}{d x} {(x - 1)}^{2}) + (\frac{d}{d x} x) ({(x - 1)}^{2})$ $(d/dyy^2)(dy/dx) = (x)(d/dx(x-1)^2) + (d/dx x)((x-1)^2)$

$:. 2ydy/dx = x(2(x-1)(1)) + (1)(x-1)^2$

$:. 2ydy/dx = 2x(x-1) + (x-1)^2$

$:. 2ydy/dx = (x-1)(2x+(x-1))$

$:. 2ydy/dx = (x-1)(3x-1)$

$:. \ \ \ dy/dx = ((x-1)(3x-1))/(2y)$

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Find $\frac{d y}{d x}$ $dy/dx$ if $y^{2} = x {(x - 1)}^{2}$ $y^2=x(x-1)^2$ ?

1 Answer

Explanation:

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