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

The answer to (\partial)/(\partial x)(6x^2+3y^2+xy) is 12x+y

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

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

12 x + y

Solution steps

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

Treat

y

as a constant

Apply the Sum/Difference Rule:

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

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

\frac{\partial}{\partial x} (6 x^{2}) = 12 x

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

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

= 12 x + 0 + y

Simplify

= 12 x + y

7 x + 1 (9.8)

f^{^{'}} (x) = e^{x^{2}}

\int \frac{1}{3 x + 4} d x

x \to 0 lim (\frac{2}{x ^{3}})

x^{2} \frac{d y}{d x} + y^{2} - x y = 0

What is (\partial)/(\partial x)(6x^2+3y^2+xy) ?
The answer to (\partial)/(\partial x)(6x^2+3y^2+xy) is 12x+y