What is (\partial)/(\partial x)(x^2-5xy+3y^2+x-4y) ?

The answer to (\partial)/(\partial x)(x^2-5xy+3y^2+x-4y) is 2x+1-5y

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(\partial)/(\partial x)(x^2-5xy+3y^2+x-4y)

\frac{\partial}{\partial x} (x^{2} - 5 x y + 3 y^{2} + x - 4 y)

2 x + 1 - 5 y

Solution steps

\frac{\partial}{\partial x} (x^{2} - 5 x y + 3 y^{2} + x - 4 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} (5 x y) + \frac{\partial}{\partial x} (3 y^{2}) + \frac{\partial}{\partial x} (x) - \frac{\partial}{\partial x} (4 y)

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

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

\frac{\partial}{\partial x} (3 y^{2}) = 0

\frac{\partial}{\partial x} (x) = 1

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

= 2 x - 5 y + 0 + 1 - 0

Simplify

= 2 x + 1 - 5 y

\int (x x^{4}) d x

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

y^{^{'''}} - 5 y^{^{''}} + 8 y^{^{'}} + 14 y = 0

\int \frac{12 x ^{3} + 6 x ^{2}}{3 x ^{4} + 2 x ^{3} - 5} d x

\frac{\partial}{\partial t} (1 + t)

What is (\partial)/(\partial x)(x^2-5xy+3y^2+x-4y) ?
The answer to (\partial)/(\partial x)(x^2-5xy+3y^2+x-4y) is 2x+1-5y