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

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

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

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

2 y

Solution steps

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

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} (x^{2}) = 0

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

= 0 + 2 y

Simplify

= 2 y

n = 1 \sum \infty n^{2} (- 1)^{n}

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

f (x) = \frac{e ^{x}}{4 x ^{2} + 1}

\int \frac{1}{x ( a ^{2} - x ^{2} )} d x

\int \frac{1}{4 + 9 x ^{2}} d x