What is (\partial)/(\partial x)(yx^3-xy^2+xy+13y) ?

The answer to (\partial)/(\partial x)(yx^3-xy^2+xy+13y) is -y^2+3yx^2+y

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(\partial)/(\partial x)(yx^3-xy^2+xy+13y)

\frac{\partial}{\partial x} (y x^{3} - x y^{2} + x y + 13 y)

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

Solution steps

\frac{\partial}{\partial x} (y x^{3} - x y^{2} + x y + 13 y)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (y x^{3}) - \frac{\partial}{\partial x} (x y^{2}) + \frac{\partial}{\partial x} (x y) + \frac{\partial}{\partial x} (13 y)

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

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

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

\frac{\partial}{\partial x} (13 y) = 0

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

Simplify

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

\int \frac{x}{x ^{2} + y ^{2}} d y

f (x) = 9 e^{x}, at x = 7

\int_{1}^{12} \frac{1}{x - 7} d x

\frac{\partial}{\partial y} (x^{2} + 6 x y + 4 y^{2})

\int x^{2} + 2 x + 1 d x

What is (\partial)/(\partial x)(yx^3-xy^2+xy+13y) ?
The answer to (\partial)/(\partial x)(yx^3-xy^2+xy+13y) is -y^2+3yx^2+y