What is the derivative of y= 1/(\sqrt[5]{2x-1)} ?

The derivative of y= 1/(\sqrt[5]{2x-1)} is -2/(5(2x-1)^{6/5)}

What is the first derivative of y= 1/(\sqrt[5]{2x-1)} ?

The first derivative of y= 1/(\sqrt[5]{2x-1)} is -2/(5(2x-1)^{6/5)}

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derivative of

- \frac{2}{5 ( 2 x - 1 ) ^{\frac{6}{5}}}

Solution steps

Apply exponent rule:

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

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

Apply the chain rule:

- \frac{1}{5 ( 2 x - 1 ) ^{\frac{6}{5}}} \frac{d}{d x} (2 x - 1)

= - \frac{1}{5 ( 2 x - 1 ) ^{\frac{6}{5}}} \frac{d}{d x} (2 x - 1)

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

= - \frac{1}{5 ( 2 x - 1 ) ^{\frac{6}{5}}} \cdot 2

Simplify

- \frac{1}{5 ( 2 x - 1 ) ^{\frac{6}{5}}} \cdot 2 : - \frac{2}{5 ( 2 x - 1 ) ^{\frac{6}{5}}}

= - \frac{2}{5 ( 2 x - 1 ) ^{\frac{6}{5}}}

\int 4 3 - 2 x - x^{2} d x

\int (x + 1) sin (n x) d x

\frac{d}{d x} (e^{- (\frac{x}{200})^{3.2}})

\int tan (4 x + 2) d x

\frac{t}{( 1 - t ^{2} ) ^{\frac{3}{2}}}

What is the derivative of y= 1/(\sqrt[5]{2x-1)} ?
The derivative of y= 1/(\sqrt[5]{2x-1)} is -2/(5(2x-1)^{6/5)}
What is the first derivative of y= 1/(\sqrt[5]{2x-1)} ?
The first derivative of y= 1/(\sqrt[5]{2x-1)} is -2/(5(2x-1)^{6/5)}