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

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

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(\partial)/(\partial y)(2x+y+1)

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

1

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

= 0 + 1 + 0

Simplify

= 1

\int \frac{4}{cos ^{2} ( 2 x )} d x

\frac{x ^{2}}{6 + x}

\int \frac{x}{x ^{2} - 2 x + 2} d x

x \to 7 lim (sin (\frac{π x}{6}))

x^{2} y^{^{'}} + 2 x y = y^{3}

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