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 \frac{n}{n ^{3} + 2}

n = 0 \sum \infty \frac{1}{2 ^{n + 2}}

\int \frac{x ^{2}}{( x ^{2} + 1 ) ^{\frac{3}{2}}} d x

\frac{d y}{d x} = - \frac{2}{x}

\int \frac{x}{( x ^{2} + 4 ) ^{3}} d x