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

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

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

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

2 x + 3 y

Solution steps

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

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

= 2 x + 3 y + 0 - 0

Simplify

= 2 x + 3 y

f (x) = - 3 x^{3} - 2 x^{2} - 1

\int_{1}^{32} x^{- \frac{9}{5}} d x

(- 9 x^{2} + 4)^{2}

\int \frac{2 x ^{4} + 3 x}{2 x ^{2}} d x

\int \frac{1}{1 + t} d t

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