What is the d/(dy)((y^3)/8) ?

The d/(dy)((y^3)/8) is (3y^2)/8

What is the first d/(dy)((y^3)/8) ?

The first d/(dy)((y^3)/8) is (3y^2)/8

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\frac{d}{d y} (\frac{y ^{3}}{8})

\frac{3 y ^{2}}{8}

Solution steps

\frac{d}{d y} (\frac{y ^{3}}{8})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{8} \frac{d}{d y} (y^{3})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= \frac{1}{8} \cdot 3 y^{3 - 1}

Simplify

\frac{1}{8} \cdot 3 y^{3 - 1} : \frac{3 y ^{2}}{8}

= \frac{3 y ^{2}}{8}

\frac{d}{d t} (sin (2 π t))

y = x^{4} + 8 x^{2} - x, (1, 8)

\int x + 2 - \frac{1}{x + 1} d x

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

n = 1 \sum \infty \frac{1}{3} (\frac{e}{2})^{n}