How do you find the derivative of $y^{3} = x^{2} - 1$ $y^3 = x^2 -1$ at P(2,1)?

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How do you find the derivative of $y^{3} = x^{2} - 1$ $y^3 = x^2 -1$ at P(2,1)?

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Answer 1 · 2016-10-13T21:10:38

The point $(2, 1)$ $(2,1)$ is not on the curve. However, the derivative at any point is:
$\frac{d y}{d x} = \frac{2}{3} \frac{x}{y^{2}}; x \neq \pm 1$ $dy/dx = 2/3x/(y^2); x ne +-1$ because x equal to plus or minus one will cause y to become zero and that is not allowed.

Explanation:

Let's check whether the point $(2, 1)$ $(2, 1)$ is on the curve by substituting 2 for x in the equation:

$y^{3} = 2^{2} - 1$ $y^3 = 2^2 - 1$

$y^{3} = 4 - 1$ $y^3 = 4 - 1$

$y^{3} = 3$ $y^3 = 3$

$y = \sqrt[3]{3}$ $y = root(3)3$

Let's find the derivative at any point:

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

$\frac{d y}{d x} = \frac{2}{3} \frac{x}{y^{2}}; x \neq \pm 1$ $dy/dx = 2/3x/(y^2); x ne +-1$

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How do you find the derivative of $y^{3} = x^{2} - 1$ $y^3 = x^2 -1$ at P(2,1)?

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How do you find the derivative of $y^{3} = x^{2} - 1$ $y^3 = x^2 -1$ at P(2,1)?