What is (\partial)/(\partial x)(x^2+3xy+y^2+1) ?

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

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(\partial)/(\partial x)(x^2+3xy+y^2+1)

\frac{\partial}{\partial x} (x^{2} + 3 x y + y^{2} + 1)

2 x + 3 y

Solution steps

\frac{\partial}{\partial x} (x^{2} + 3 x y + y^{2} + 1)

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} (y^{2}) + \frac{\partial}{\partial x} (1)

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

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

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

\frac{\partial}{\partial x} (1) = 0

= 2 x + 3 y + 0 + 0

Simplify

= 2 x + 3 y

\int \frac{1}{3 - 2 y} d y

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

3 e^{- x}

\frac{\partial}{\partial y} (sin (\frac{x}{1 + y}))

n = 0 \sum \infty \frac{1}{n ^{6}}

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