What is (\partial)/(\partial x)(x^2-3xy+6y^2+x-2y) ?

The answer to (\partial)/(\partial x)(x^2-3xy+6y^2+x-2y) is 2x+1-3y

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(\partial)/(\partial x)(x^2-3xy+6y^2+x-2y)

\frac{\partial}{\partial x} (x^{2} - 3 x y + 6 y^{2} + x - 2 y)

2 x + 1 - 3 y

Solution steps

\frac{\partial}{\partial x} (x^{2} - 3 x y + 6 y^{2} + x - 2 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} (3 x y) + \frac{\partial}{\partial x} (6 y^{2}) + \frac{\partial}{\partial x} (x) - \frac{\partial}{\partial x} (2 y)

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

\frac{\partial}{\partial x} (3 x y) = 3 y

\frac{\partial}{\partial x} (6 y^{2}) = 0

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

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

= 2 x - 3 y + 0 + 1 - 0

Simplify

= 2 x + 1 - 3 y

\frac{\partial}{\partial x} (y \cdot ln (x^{2} + 5 y + 40))

\frac{d}{d t} (e^{\frac{t}{2}})

x \to \infty + lim (5 x + 6)

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

x \to 2 - lim (1 - x^{2})

What is (\partial)/(\partial x)(x^2-3xy+6y^2+x-2y) ?
The answer to (\partial)/(\partial x)(x^2-3xy+6y^2+x-2y) is 2x+1-3y