What is the derivative of f(x)=(9\sqrt[3]{x^5})/5 ?

The derivative of f(x)=(9\sqrt[3]{x^5})/5 is (3x^4)/((x^5)^{2/3)}

What is the first derivative of f(x)=(9\sqrt[3]{x^5})/5 ?

The first derivative of f(x)=(9\sqrt[3]{x^5})/5 is (3x^4)/((x^5)^{2/3)}

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derivative of

\frac{3 x ^{4}}{( x ^{5} ) ^{\frac{2}{3}}}

Solution steps

Take the constant out:

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

Apply the chain rule:

\frac{1}{3 ( x ^{5} ) ^{\frac{2}{3}}} \frac{d}{d x} (x^{5})

= \frac{1}{3 ( x ^{5} ) ^{\frac{2}{3}}} \frac{d}{d x} (x^{5})

\frac{d}{d x} (x^{5}) = 5 x^{4}

= \frac{9}{5} \cdot \frac{1}{3 ( x ^{5} ) ^{\frac{2}{3}}} \cdot 5 x^{4}

Simplify

\frac{9}{5} \cdot \frac{1}{3 ( x ^{5} ) ^{\frac{2}{3}}} \cdot 5 x^{4} : \frac{3 x ^{4}}{( x ^{5} ) ^{\frac{2}{3}}}

= \frac{3 x ^{4}}{( x ^{5} ) ^{\frac{2}{3}}}

\int e^{(x + y)}

x y^{2} y^{^{'}} = x + 8

\int \frac{2 x}{4 x ^{4} - 9} d x

\int (6 x + 1) (6 x^{2} + 2 x)^{\frac{3}{4}} d x

\frac{d}{d x} (\frac{5}{x + 8})

What is the derivative of f(x)=(9\sqrt[3]{x^5})/5 ?
The derivative of f(x)=(9\sqrt[3]{x^5})/5 is (3x^4)/((x^5)^{2/3)}
What is the first derivative of f(x)=(9\sqrt[3]{x^5})/5 ?
The first derivative of f(x)=(9\sqrt[3]{x^5})/5 is (3x^4)/((x^5)^{2/3)}