What is the slope of the tangent line of $\frac{x^{2}}{x - y^{3}} = C$ $x^2/(x-y^3) = C$ , where C is an arbitrary constant, at $(- 2, 1)$ $(-2,1)$ ?

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What is the slope of the tangent line of $\frac{x^{2}}{x - y^{3}} = C$ $x^2/(x-y^3) = C$ , where C is an arbitrary constant, at $(- 2, 1)$ $(-2,1)$ ?

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Answer 1 · 2016-01-02T08:02:11

$- \frac{3}{2}$ $-3/2$

Explanation:

Find the implicit derivative of the function.

Use the product rule.

$\frac{(x - y^{3}) \frac{d}{d x} (x^{2}) - x^{2} \frac{d}{d x} (x - y^{3})}{{(x - y^{3})}^{2}} = 0$ $((x-y^3)d/dx(x^2)-x^2d/dx(x-y^3))/(x-y^3)^2=0$

Remember that differentiating a $y$ $y$ term will spit out a $\frac{d y}{d x}$ $dy/dx$ term thanks to the chain rule.

$\frac{2 x (x - y^{3}) - x^{2} (1 - 3 y^{2} \frac{d y}{d x})}{{(x - y^{3})}^{2}} = 0$ $(2x(x-y^3)-x^2(1-3y^2dy/dx))/(x-y^3)^2=0$

Plug in $(- 2, 1)$ $(-2,1)$ for $(x, y)$ $(x,y)$ and solve for $\frac{d y}{d x}$ $dy/dx$ .

$\frac{2 (- 2) (- 2 - 1^{3}) - {(- 2)}^{2} (1 - 3 {(1)}^{2} \frac{d y}{d x})}{{(- 2 - 1^{3})}^{2}} = 0$ $(2(-2)(-2-1^3)-(-2)^2(1-3(1)^2dy/dx))/(-2-1^3)^2=0$

Notice that the denominator can be cancelled out. Continue simplification of numerator.

$(- 4) (- 3) - (4) (- 2) \frac{d y}{d x} = 0$ $(-4)(-3)-(4)(-2)dy/dx=0$

$12 + 8 \frac{d y}{d x} = 0$ $12+8dy/dx=0$

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

The slope of the tangent line is $- \frac{3}{2}$ $-3/2$ .

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What is the slope of the tangent line of $\frac{x^{2}}{x - y^{3}} = C$ $x^2/(x-y^3) = C$ , where C is an arbitrary constant, at $(- 2, 1)$ $(-2,1)$ ?

1 Answer

Explanation:

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What is the slope of the tangent line of x^2/(x-y^3) = C x2x−y3=C x^2/(x-y^3) = C , where C is an arbitrary constant, at (-2,1)(−2,1)(-2,1)?

1 Answer

Explanation:

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What is the slope of the tangent line of $\frac{x^{2}}{x - y^{3}} = C$ $x^2/(x-y^3) = C$ , where C is an arbitrary constant, at $(- 2, 1)$ $(-2,1)$ ?