What is (\partial)/(\partial x)(250-4x-7y+0.2x^2+0.1y^2) ?

The answer to (\partial)/(\partial x)(250-4x-7y+0.2x^2+0.1y^2) is 0.4x-4

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(\partial)/(\partial x)(250-4x-7y+0.2x^2+0.1y^2)

\frac{\partial}{\partial x} (250 - 4 x - 7 y + 0.2 x^{2} + 0.1 y^{2})

0.4 x - 4

Solution steps

\frac{\partial}{\partial x} (250 - 4 x - 7 y + 0.2 x^{2} + 0.1 y^{2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (250) - \frac{\partial}{\partial x} (4 x) - \frac{\partial}{\partial x} (7 y) + \frac{\partial}{\partial x} (0.2 x^{2}) + \frac{\partial}{\partial x} (0.1 y^{2})

\frac{\partial}{\partial x} (250) = 0

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

\frac{\partial}{\partial x} (7 y) = 0

\frac{\partial}{\partial x} (0.2 x^{2}) = 0.4 x

\frac{\partial}{\partial x} (0.1 y^{2}) = 0

= 0 - 4 - 0 + 0.4 x + 0

Simplify

= 0.4 x - 4

\int \frac{3}{2} x + 1 d x

sin (6 (t - \frac{3 π}{6}) + 3 π)

\frac{d y}{d x} = \frac{e ^{x - y}}{1 + e ^{x}}

\int \frac{x ^{2}}{( x ^{2} - 1 ) ^{\frac{3}{2}}} d x

f (x) = x^{2} - 5 x, at x = - 2

What is (\partial)/(\partial x)(250-4x-7y+0.2x^2+0.1y^2) ?
The answer to (\partial)/(\partial x)(250-4x-7y+0.2x^2+0.1y^2) is 0.4x-4