What is the derivative of x^3+y^3+(1-x-y)^3 ?

The derivative of x^3+y^3+(1-x-y)^3 is -3y^2+6y-6xy+6x-3

What is the first derivative of x^3+y^3+(1-x-y)^3 ?

The first derivative of x^3+y^3+(1-x-y)^3 is -3y^2+6y-6xy+6x-3

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derivative of

x^{3} + y^{3} + (1 - x - y)^{3}

- 3 y^{2} + 6 y - 6 x y + 6 x - 3

Solution steps

\frac{d}{d x} (x^{3} + y^{3} + (1 - x - y)^{3})

Simplify

x^{3} + y^{3} + (1 - x - y)^{3} : - 3 x^{2} y + 3 x^{2} - 3 x y^{2} + 6 x y - 3 x + 3 y^{2} - 3 y + 1

= \frac{d}{d x} (- 3 x^{2} y + 3 x^{2} - 3 x y^{2} + 6 x y - 3 x + 3 y^{2} - 3 y + 1)

Treat

y

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= - \frac{d}{d x} (3 x^{2} y) + \frac{d}{d x} (3 x^{2}) - \frac{d}{d x} (3 x y^{2}) + \frac{d}{d x} (6 x y) - \frac{d}{d x} (3 x) + \frac{d}{d x} (3 y^{2}) - \frac{d}{d x} (3 y) + \frac{d}{d x} (1)

\frac{d}{d x} (3 x^{2} y) = 6 y x

\frac{d}{d x} (3 x^{2}) = 6 x

\frac{d}{d x} (3 x y^{2}) = 3 y^{2}

\frac{d}{d x} (6 x y) = 6 y

\frac{d}{d x} (3 x) = 3

\frac{d}{d x} (3 y^{2}) = 0

\frac{d}{d x} (3 y) = 0

\frac{d}{d x} (1) = 0

= - 6 y x + 6 x - 3 y^{2} + 6 y - 3 + 0 - 0 + 0

Simplify

= - 3 y^{2} + 6 y - 6 x y + 6 x - 3

x^{\frac{1}{2}} lo g_{10} (x)

\frac{( 1 - lo g _{10} ( x ) )}{( x ^{2} )}

x (ln (x))^{2}

(x^{10})^{x}

\frac{8 x ^{2}}{3}

What is the derivative of x^3+y^3+(1-x-y)^3 ?
The derivative of x^3+y^3+(1-x-y)^3 is -3y^2+6y-6xy+6x-3
What is the first derivative of x^3+y^3+(1-x-y)^3 ?
The first derivative of x^3+y^3+(1-x-y)^3 is -3y^2+6y-6xy+6x-3