What is the d/(dy)(1/(4y^2)) ?

The d/(dy)(1/(4y^2)) is -1/(2y^3)

What is the first d/(dy)(1/(4y^2)) ?

The first d/(dy)(1/(4y^2)) is -1/(2y^3)

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\frac{d}{d y} (\frac{1}{4 y ^{2}})

- \frac{1}{2 y ^{3}}

Solution steps

\frac{d}{d y} (\frac{1}{4 y ^{2}})

Take the constant out:

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

= \frac{1}{4} \frac{d}{d y} (\frac{1}{y ^{2}})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= \frac{1}{4} \frac{d}{d y} (y^{- 2})

Apply the Power Rule:

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

= \frac{1}{4} (- 2 y^{- 2 - 1})

Simplify

\frac{1}{4} (- 2 y^{- 2 - 1}) : - \frac{1}{2 y ^{3}}

= - \frac{1}{2 y ^{3}}

\frac{\partial}{\partial y} (\frac{3 x y}{x ^{2} + y ^{2}})

2 y \frac{d y}{d x} = x (y^{2} + 1)

cos (x) d y + (y sin (x) - cos^{4} (x)) d x = 0

f (x) = x + 2 sin (x)

\frac{\partial}{\partial x} ((x^{2} + x y)^{3})