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

The d/(dy)(x/(2y^4)) is -(2x)/(y^5)

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

The first d/(dy)(x/(2y^4)) is -(2x)/(y^5)

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

- \frac{2 x}{y ^{5}}

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the Power Rule:

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

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

Simplify

\frac{x}{2} (- 4 y^{- 4 - 1}) : - \frac{2 x}{y ^{5}}

= - \frac{2 x}{y ^{5}}

\frac{\partial}{\partial z} (x cos (y))

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

2 x y^{^{'}} - 6 y = x^{- 4}

x \to (\frac{π}{2}) - lim (sec (x))

\frac{d}{d x} ((cot (x) - csc (x))^{- 1})