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

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

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

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

2 x + y - 2

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

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

\frac{\partial}{\partial x} (5 y) = 0

= 2 x + y + 0 - 2 - 0

Simplify

= 2 x + y - 2

f (x) = e^{5 x^{2} - x}

\frac{d}{d x} (\frac{3 x ^{5} - 7 x ^{2} - 4}{x ^{2}})

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

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

\frac{3}{x + 4}

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