What is the derivative of y= 1/(\sqrt[3]{x-1)} ?

The derivative of y= 1/(\sqrt[3]{x-1)} is -1/(3(x-1)^{4/3)}

What is the first derivative of y= 1/(\sqrt[3]{x-1)} ?

The first derivative of y= 1/(\sqrt[3]{x-1)} is -1/(3(x-1)^{4/3)}

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derivative of

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

Solution steps

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

\int (cos^{3} (x) sin (x)) d x

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

(x^{2} + y^{2}) d x + (2 x y) d y = 0

y^{^{'}} = \frac{x ^{2} - y ^{2}}{x y}, y (2) = 5

\frac{\partial}{\partial y} (e^{x y z^{4}} y z^{4})

What is the derivative of y= 1/(\sqrt[3]{x-1)} ?
The derivative of y= 1/(\sqrt[3]{x-1)} is -1/(3(x-1)^{4/3)}
What is the first derivative of y= 1/(\sqrt[3]{x-1)} ?
The first derivative of y= 1/(\sqrt[3]{x-1)} is -1/(3(x-1)^{4/3)}