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

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

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

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

2 y

Solution steps

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

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

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

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

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

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

= 0 + 2 y + 0

Simplify

= 2 y

\int ln (cos (w)) tan (w) d w

(2 x y) d x + (1 - x^{2}) d y = 0

x \to \infty lim (\frac{ln ( x ) + x ^{3}}{e ^{- x}})

n = 0 \sum \infty \frac{9}{3 ^{n - 1}}

x \to 2 lim (x^{2} - 43 - x + 7)

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