What is (\partial)/(\partial x)(5xy+2x-x^2-4y^2) ?

The answer to (\partial)/(\partial x)(5xy+2x-x^2-4y^2) is 5y+2-2x

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(\partial)/(\partial x)(5xy+2x-x^2-4y^2)

\frac{\partial}{\partial x} (5 x y + 2 x - x^{2} - 4 y^{2})

5 y + 2 - 2 x

Solution steps

\frac{\partial}{\partial x} (5 x y + 2 x - x^{2} - 4 y^{2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

= \frac{\partial}{\partial x} (5 x y) + \frac{\partial}{\partial x} (2 x) - \frac{\partial}{\partial x} (x^{2}) - \frac{\partial}{\partial x} (4 y^{2})

\frac{\partial}{\partial x} (5 x y) = 5 y

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

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

\frac{\partial}{\partial x} (4 y^{2}) = 0

= 5 y + 2 - 2 x - 0

Simplify

= 5 y + 2 - 2 x

\int 8 tan^{7} (x) sec^{2} (x) d x

\frac{d y}{d x} = 8 - y

\frac{d}{d x} (\frac{x ^{3}}{3} - 4 x)

f (x) = \frac{2 x - 3}{x ^{2} + 2 x}

2 x^{3} - e^{x y} = x \cdot y^{4}

What is (\partial)/(\partial x)(5xy+2x-x^2-4y^2) ?
The answer to (\partial)/(\partial x)(5xy+2x-x^2-4y^2) is 5y+2-2x