What is derivative of x^3-xy+y^3=1 ?

The answer to derivative of x^3-xy+y^3=1 is (dy)/(dx)=(-3x^2+y)/(-x+3y^2)

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derivative of x^3-xy+y^3=1

derivative of

x^{3} - x y + y^{3} = 1

\frac{d y}{d x} = \frac{- 3 x ^{2} + y}{- x + 3 y ^{2}}

Solution steps

x^{3} - x y + y^{3} = 1

Treat

y

y (x)

Differentiate both sides:

3 x^{2} - y - x \frac{d y}{d x} + 3 y^{2} \frac{d y}{d x} = 0

Isolate

\frac{d y}{d x} : \frac{d y}{d x} = \frac{- 3 x ^{2} + y}{- x + 3 y ^{2}}

\frac{d y}{d x} = \frac{- 3 x ^{2} + y}{- x + 3 y ^{2}}

(17, - 11), (- 14, - 16)

(1, 1)

y = - x

y = 3 x - 2, (3, 7)

f (x) = sec^{2} (x)

What is derivative of x^3-xy+y^3=1 ?
The answer to derivative of x^3-xy+y^3=1 is (dy)/(dx)=(-3x^2+y)/(-x+3y^2)