What is (\partial)/(\partial x)(50x-0.5x^2+75y-y^2) ?

The answer to (\partial)/(\partial x)(50x-0.5x^2+75y-y^2) is -x+50

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(\partial)/(\partial x)(50x-0.5x^2+75y-y^2)

\frac{\partial}{\partial x} (50 x - 0.5 x^{2} + 75 y - y^{2})

- x + 50

Solution steps

\frac{\partial}{\partial x} (50 x - 0.5 x^{2} + 75 y - y^{2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (50 x) - \frac{\partial}{\partial x} (0.5 x^{2}) + \frac{\partial}{\partial x} (75 y) - \frac{\partial}{\partial x} (y^{2})

\frac{\partial}{\partial x} (50 x) = 50

\frac{\partial}{\partial x} (0.5 x^{2}) = x

\frac{\partial}{\partial x} (75 y) = 0

\frac{\partial}{\partial x} (y^{2}) = 0

= 50 - x + 0 - 0

Simplify

= - x + 50

\frac{\partial}{\partial x} (- ln (x^{2} + 1))

n = 1 \sum \infty (x + 6)^{n}

\int e^{- x^{3}} d x

\int x (x^{2} + 4)^{\frac{3}{2}} d x

\frac{\partial}{\partial x} (4)

What is (\partial)/(\partial x)(50x-0.5x^2+75y-y^2) ?
The answer to (\partial)/(\partial x)(50x-0.5x^2+75y-y^2) is -x+50