What is (\partial)/(\partial x)(x^3+y^3-6xy) ?

The answer to (\partial)/(\partial x)(x^3+y^3-6xy) is 3x^2-6y

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(\partial)/(\partial x)(x^3+y^3-6xy)

\frac{\partial}{\partial x} (x^{3} + y^{3} - 6 x y)

3 x^{2} - 6 y

Solution steps

\frac{\partial}{\partial x} (x^{3} + y^{3} - 6 x y)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (x^{3}) + \frac{\partial}{\partial x} (y^{3}) - \frac{\partial}{\partial x} (6 x y)

\frac{\partial}{\partial x} (x^{3}) = 3 x^{2}

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

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

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

Simplify

= 3 x^{2} - 6 y

\int sec^{2} (3 x - 4) d x

n = 0 \sum \infty n

\int 1 - cos^{2} (2 x) d x

R (x) = 71 x

\int \frac{3}{5 y y ^{2} - 1} d y

What is (\partial)/(\partial x)(x^3+y^3-6xy) ?
The answer to (\partial)/(\partial x)(x^3+y^3-6xy) is 3x^2-6y