How do you find $\frac{d y}{d x}$ $dy/dx$ by implicit differentiation of $x^{3} + y^{3} = 8$ $x^3+y^3=8$ ?

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Answer 1 · 2017-02-13T01:18:37

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

We have:

$x^{3} + y^{3} = 8$ $x^3+y^3=8$

Differentiate wrt $x$ $x$ ; then Apply the chain rule (implicit differentiation):

$3 x^{2} + \frac{d}{d x} y^{3} = 0$ $3x^2 + d/dxy^3=0$
$:. 3x^2 + dy/dxd/dyy^3=0$
$:. 3x^2 + dy/dx(3y^2)=0$
$:. x^2 + y^2dy/dx=0$
$:. dy/dx=-x^2/y^2$

How do you find dy/dxdydxdy/dx by implicit differentiation of x^3+y^3=8x3+y3=8x^3+y^3=8?