What is (\partial)/(\partial y)((9x^2)/(y^{1/3)}) ?

The answer to (\partial)/(\partial y)((9x^2)/(y^{1/3)}) is -(3x^2)/(y^{4/3)}

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(\partial)/(\partial y)((9x^2)/(y^{1/3)})

\frac{\partial}{\partial y} (\frac{9 x ^{2}}{y ^{\frac{1}{3}}})

- \frac{3 x ^{2}}{y ^{\frac{4}{3}}}

Solution steps

\frac{\partial}{\partial y} (\frac{9 x ^{2}}{y ^{\frac{1}{3}}})

Treat

x

as a constant

Take the constant out:

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

= 9 x^{2} \frac{\partial}{\partial y} (\frac{1}{y ^{\frac{1}{3}}})

Apply exponent rule:

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

= 9 x^{2} \frac{\partial}{\partial y} (y^{- \frac{1}{3}})

Apply the Power Rule:

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

= 9 x^{2} (- \frac{1}{3} y^{- \frac{1}{3} - 1})

Simplify

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

= - \frac{3 x ^{2}}{y ^{\frac{4}{3}}}

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

f (x) = \frac{x ^{2}}{1 + 8 x}

- 5 y

3 (x^{3} - x), 0

f (x) = 4 x^{2} + 12 x + 1, at x = - 2

What is (\partial)/(\partial y)((9x^2)/(y^{1/3)}) ?
The answer to (\partial)/(\partial y)((9x^2)/(y^{1/3)}) is -(3x^2)/(y^{4/3)}