What is (\partial)/(\partial l)(9l^{2/3}k^{1/3}) ?

The answer to (\partial)/(\partial l)(9l^{2/3}k^{1/3}) is (6k^{1/3})/(l^{1/3)}

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(\partial)/(\partial l)(9l^{2/3}k^{1/3})

\frac{\partial}{\partial l} (9 l^{\frac{2}{3}} k^{\frac{1}{3}})

\frac{6 k ^{\frac{1}{3}}}{l ^{\frac{1}{3}}}

Solution steps

\frac{\partial}{\partial l} (9 l^{\frac{2}{3}} k^{\frac{1}{3}})

Take the constant out:

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

= 9 k^{\frac{1}{3}} \frac{\partial}{\partial l} (l^{\frac{2}{3}})

Apply the Power Rule:

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

= 9 k^{\frac{1}{3}} \frac{2}{3} l^{\frac{2}{3} - 1}

Simplify

9 k^{\frac{1}{3}} \frac{2}{3} l^{\frac{2}{3} - 1} : \frac{6 k ^{\frac{1}{3}}}{l ^{\frac{1}{3}}}

= \frac{6 k ^{\frac{1}{3}}}{l ^{\frac{1}{3}}}

y^{^{''}} + 9 y = 24 sin (3 t)

\int \frac{2 ^{x + 1}}{1 0 ^{x}} d x

\int \frac{1 + sin ( x )}{cos ^{2} ( x )} d x

3 ln (4 - x^{2})

y^{^{'}} = \frac{ln ( 2 )}{7} y - 20, y (0) = 200

What is (\partial)/(\partial l)(9l^{2/3}k^{1/3}) ?
The answer to (\partial)/(\partial l)(9l^{2/3}k^{1/3}) is (6k^{1/3})/(l^{1/3)}