What is (\partial)/(\partial x)(x^3+x^2y^2+y^3+x+y) ?

The answer to (\partial)/(\partial x)(x^3+x^2y^2+y^3+x+y) is 3x^2+2xy^2+1

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(\partial)/(\partial x)(x^3+x^2y^2+y^3+x+y)

\frac{\partial}{\partial x} (x^{3} + x^{2} y^{2} + y^{3} + x + y)

3 x^{2} + 2 x y^{2} + 1

Solution steps

\frac{\partial}{\partial x} (x^{3} + x^{2} y^{2} + y^{3} + 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} (x^{2} y^{2}) + \frac{\partial}{\partial x} (y^{3}) + \frac{\partial}{\partial x} (x) + \frac{\partial}{\partial x} (y)

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

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

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

\frac{\partial}{\partial x} (x) = 1

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

= 3 x^{2} + 2 y^{2} x + 0 + 1 + 0

Simplify

= 3 x^{2} + 2 x y^{2} + 1

\frac{\partial}{\partial x} (ln (x + 2 y + 9 z))

- (1 - 4 \cdot z)

\int sin (3 x) cos (2 x) d x

\int \frac{sec ( \frac{1}{x} ) tan ( \frac{1}{x} )}{x ^{2}} d x

\frac{d}{d x} (\frac{94 x}{99 x - 94})

What is (\partial)/(\partial x)(x^3+x^2y^2+y^3+x+y) ?
The answer to (\partial)/(\partial x)(x^3+x^2y^2+y^3+x+y) is 3x^2+2xy^2+1