What is the derivative of \sqrt[7]{2x^2-1} ?

The derivative of \sqrt[7]{2x^2-1} is (4x)/(7(2x^2-1)^{6/7)}

What is the first derivative of \sqrt[7]{2x^2-1} ?

The first derivative of \sqrt[7]{2x^2-1} is (4x)/(7(2x^2-1)^{6/7)}

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\frac{4 x}{7 ( 2 x ^{2} - 1 ) ^{\frac{6}{7}}}

Solution steps

Apply the chain rule:

\frac{1}{7 ( 2 x ^{2} - 1 ) ^{\frac{6}{7}}} \frac{d}{d x} (2 x^{2} - 1)

= \frac{1}{7 ( 2 x ^{2} - 1 ) ^{\frac{6}{7}}} \frac{d}{d x} (2 x^{2} - 1)

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

= \frac{1}{7 ( 2 x ^{2} - 1 ) ^{\frac{6}{7}}} \cdot 4 x

Simplify

\frac{1}{7 ( 2 x ^{2} - 1 ) ^{\frac{6}{7}}} \cdot 4 x : \frac{4 x}{7 ( 2 x ^{2} - 1 ) ^{\frac{6}{7}}}

= \frac{4 x}{7 ( 2 x ^{2} - 1 ) ^{\frac{6}{7}}}

\int sin (3 x) + cos (5 x) d x

y = e^{- \frac{7}{t ^{2}}}

y = - 3 x + 2

\frac{3 s + 3}{s ^{2} - 9}

- x^{4} - x^{2}

What is the derivative of \sqrt[7]{2x^2-1} ?
The derivative of \sqrt[7]{2x^2-1} is (4x)/(7(2x^2-1)^{6/7)}
What is the first derivative of \sqrt[7]{2x^2-1} ?
The first derivative of \sqrt[7]{2x^2-1} is (4x)/(7(2x^2-1)^{6/7)}