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 1 + e^{x} d x

\frac{d}{d x} (ln (x^{8} + 19))

\frac{d}{d x} (2 cos (x^{3}) - \frac{2}{x})

\frac{d}{d t} (t^{3})

f (x) = x^{2} + 3 x + 1, (- 1, - 1)

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