What is (\partial)/(\partial x)(x^2y+xy^2+20) ?

The answer to (\partial)/(\partial x)(x^2y+xy^2+20) is y^2+2xy

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(\partial)/(\partial x)(x^2y+xy^2+20)

\frac{\partial}{\partial x} (x^{2} y + x y^{2} + 20)

y^{2} + 2 x y

Solution steps

\frac{\partial}{\partial x} (x^{2} y + x y^{2} + 20)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (x^{2} y) + \frac{\partial}{\partial x} (x y^{2}) + \frac{\partial}{\partial x} (20)

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

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

\frac{\partial}{\partial x} (20) = 0

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

Simplify

= y^{2} + 2 x y

\frac{d}{d x} (x^{0.7} y^{0.3})

\frac{d}{d y} (3 cos (3 x) (x - y) + sin (3 x))

\frac{d}{d x} (\frac{4 x ^{4} + 7}{x ^{2}})

y^{^{'}} = 0.06 y - 0.0006 y^{2}

n = 1 \sum \infty \frac{n ^{3}}{n ^{3}}

What is (\partial)/(\partial x)(x^2y+xy^2+20) ?
The answer to (\partial)/(\partial x)(x^2y+xy^2+20) is y^2+2xy