What is (\partial)/(\partial u)(u^{1/3}*v^{2/3}) ?

The answer to (\partial)/(\partial u)(u^{1/3}*v^{2/3}) is (v^{2/3})/(3u^{2/3)}

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(\partial)/(\partial u)(u^{1/3}*v^{2/3})

\frac{\partial}{\partial u} (u^{\frac{1}{3}} \cdot v^{\frac{2}{3}})

\frac{v ^{\frac{2}{3}}}{3 u ^{\frac{2}{3}}}

Solution steps

\frac{\partial}{\partial u} (u^{\frac{1}{3}} v^{\frac{2}{3}})

Treat

v

as a constant

Take the constant out:

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

= v^{\frac{2}{3}} \frac{\partial}{\partial u} (u^{\frac{1}{3}})

Apply the Power Rule:

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

= v^{\frac{2}{3}} \frac{1}{3} u^{\frac{1}{3} - 1}

Simplify

v^{\frac{2}{3}} \frac{1}{3} u^{\frac{1}{3} - 1} : \frac{v ^{\frac{2}{3}}}{3 u ^{\frac{2}{3}}}

= \frac{v ^{\frac{2}{3}}}{3 u ^{\frac{2}{3}}}

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

4 x^{3} - 5 7 x^{6} + π x

e^{- 4 t} cos (3 t)

\int x e^{3 x} d x

f (t) = 6 cos (t)

What is (\partial)/(\partial u)(u^{1/3}*v^{2/3}) ?
The answer to (\partial)/(\partial u)(u^{1/3}*v^{2/3}) is (v^{2/3})/(3u^{2/3)}