What is (\partial)/(\partial x)(x^2+3y^2-6y+4xy) ?

The answer to (\partial)/(\partial x)(x^2+3y^2-6y+4xy) is 2x+4y

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(\partial)/(\partial x)(x^2+3y^2-6y+4xy)

\frac{\partial}{\partial x} (x^{2} + 3 y^{2} - 6 y + 4 x y)

2 x + 4 y

Solution steps

\frac{\partial}{\partial x} (x^{2} + 3 y^{2} - 6 y + 4 x 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 y^{2}) - \frac{\partial}{\partial x} (6 y) + \frac{\partial}{\partial x} (4 x y)

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

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

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

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

= 2 x + 0 - 0 + 4 y

Simplify

= 2 x + 4 y

f (x) = - x^{2} + 2 x + 4, at x = 3

y = - 3 x^{2}

\int \frac{1}{- e ^{2 x}} d x

\frac{d}{d x} (x + \frac{1}{x + 1})

f (x) = \frac{1}{x}, at x = 1

What is (\partial)/(\partial x)(x^2+3y^2-6y+4xy) ?
The answer to (\partial)/(\partial x)(x^2+3y^2-6y+4xy) is 2x+4y