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

The derivative of 9/(\sqrt[3]{x}) is -3/(x^{4/3)}

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

The first derivative of 9/(\sqrt[3]{x}) is -3/(x^{4/3)}

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\frac{d}{d x} (\frac{9}{3 x})

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

Solution steps

\frac{d}{d x} (\frac{9}{3 x})

Take the constant out:

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

= 9 \frac{d}{d x} (\frac{1}{3 x})

Apply exponent rule:

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

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

Apply the Power Rule:

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

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

Simplify

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

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

arctan (4 ln (x))

174

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

f (x) = 6 x, at x = 6

\int \frac{t}{t ^{2} - 1} d t

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