What is (\partial)/(\partial x)(300-20x^2+30y) ?

The answer to (\partial)/(\partial x)(300-20x^2+30y) is -40x

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(\partial)/(\partial x)(300-20x^2+30y)

\frac{\partial}{\partial x} (300 - 20 x^{2} + 30 y)

- 40 x

Solution steps

\frac{\partial}{\partial x} (300 - 20 x^{2} + 30 y)

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (300) - \frac{\partial}{\partial x} (20 x^{2}) + \frac{\partial}{\partial x} (30 y)

\frac{\partial}{\partial x} (300) = 0

\frac{\partial}{\partial x} (20 x^{2}) = 40 x

\frac{\partial}{\partial x} (30 y) = 0

= 0 - 40 x + 0

Simplify

= - 40 x

x \frac{d y}{d x} + 1 y = 6 y^{2}, y (1) = - 1

n = 1 \sum \infty \frac{1}{2 n + 3}

x^{2} + 6

\frac{d}{d x} (- 49 sin (7 x))

\int 4 sin (5 x) + 5 sin (4 x) d x

What is (\partial)/(\partial x)(300-20x^2+30y) ?
The answer to (\partial)/(\partial x)(300-20x^2+30y) is -40x