What is the first d/(dy)(x^2+y^2-5) ?

The first d/(dy)(x^2+y^2-5) is 2y

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d/(dy)(x^2+y^2-5)

\frac{d}{d y} (x^{2} + y^{2} - 5)

2 y

Solution steps

\frac{d}{d y} (x^{2} + y^{2} - 5)

Treat

x

as a constant

Apply the Sum/Difference Rule:

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

= \frac{d}{d y} (x^{2}) + \frac{d}{d y} (y^{2}) - \frac{d}{d y} (5)

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

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

\frac{d}{d y} (5) = 0

= 0 + 2 y - 0

Simplify

= 2 y

g (w) = \frac{9}{8 w ^{5}} + 9 w

\int_{0}^{1} \frac{2}{5} x^{2} (x + 2) d x

\int \frac{1}{e ^{- 4 x}} d x

x \to 2 lim (\frac{x - 5}{x - 2})

x \to 0 lim (\frac{z ^{x} - 1}{x})