What is the derivative of (18)/(\sqrt[3]{x-2)} ?

The derivative of (18)/(\sqrt[3]{x-2)} is -6/((x-2)^{4/3)}

What is the first derivative of (18)/(\sqrt[3]{x-2)} ?

The first derivative of (18)/(\sqrt[3]{x-2)} is -6/((x-2)^{4/3)}

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derivative of

- \frac{6}{( x - 2 ) ^{\frac{4}{3}}}

Solution steps

Take the constant out:

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

Apply exponent rule:

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

= 18 \frac{d}{d x} ((x - 2)^{- \frac{1}{3}})

Apply the chain rule:

- \frac{1}{3 ( x - 2 ) ^{\frac{4}{3}}} \frac{d}{d x} (x - 2)

= - \frac{1}{3 ( x - 2 ) ^{\frac{4}{3}}} \frac{d}{d x} (x - 2)

\frac{d}{d x} (x - 2) = 1

= 18 (- \frac{1}{3 ( x - 2 ) ^{\frac{4}{3}}} \cdot 1)

Simplify

18 (- \frac{1}{3 ( x - 2 ) ^{\frac{4}{3}}} \cdot 1) : - \frac{6}{( x - 2 ) ^{\frac{4}{3}}}

= - \frac{6}{( x - 2 ) ^{\frac{4}{3}}}

y = x^{6} cos (x)

e^{x} (y^{3} + x y^{3} + 1) d x + 3 y^{2} (x e^{x} - 6) d y = 0

(- 7, 6), (9, - \frac{6}{7})

f (x) = 3 x^{2} - \frac{4}{x} + \frac{4}{x}

\frac{d y}{d x} = sec^{2} (x)

What is the derivative of (18)/(\sqrt[3]{x-2)} ?
The derivative of (18)/(\sqrt[3]{x-2)} is -6/((x-2)^{4/3)}
What is the first derivative of (18)/(\sqrt[3]{x-2)} ?
The first derivative of (18)/(\sqrt[3]{x-2)} is -6/((x-2)^{4/3)}