What is (\partial)/(\partial y)(x^2+y^2-25) ?

The answer to (\partial)/(\partial y)(x^2+y^2-25) is 2y

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(\partial)/(\partial y)(x^2+y^2-25)

\frac{\partial}{\partial y} (x^{2} + y^{2} - 25)

2 y

Solution steps

\frac{\partial}{\partial y} (x^{2} + y^{2} - 25)

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} (25)

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

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

\frac{\partial}{\partial y} (25) = 0

= 0 + 2 y - 0

Simplify

= 2 y

\int - 0.04 t d t

\frac{d}{d x} (- \frac{2}{9} x^{- \frac{5}{3}})

f (x) = \frac{699}{( x + 1 ) ^{0.94}}

\int_{- \frac{1}{2}}^{0} x^{4} \cdot e^{2 x + 1} d x

x \to \infty lim (\frac{x}{c})

What is (\partial)/(\partial y)(x^2+y^2-25) ?
The answer to (\partial)/(\partial y)(x^2+y^2-25) is 2y