What is (\partial)/(\partial y)((x^2)/(4y^4)) ?

The answer to (\partial)/(\partial y)((x^2)/(4y^4)) is -(x^2)/(y^5)

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(\partial)/(\partial y)((x^2)/(4y^4))

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

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

Solution steps

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

Treat

x

as a constant

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the Power Rule:

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

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

Simplify

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

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

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

\int \frac{1}{z ^{2} + 4} d z

y = x, y = 12 x, x = 16

y^{^{''}} + y = cot (x)

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

What is (\partial)/(\partial y)((x^2)/(4y^4)) ?
The answer to (\partial)/(\partial y)((x^2)/(4y^4)) is -(x^2)/(y^5)