What is the derivative of f(x)= 1/(\sqrt[3]{x^2-1)} ?

The derivative of f(x)= 1/(\sqrt[3]{x^2-1)} is -(2x)/(3(x^2-1)^{4/3)}

What is the first derivative of f(x)= 1/(\sqrt[3]{x^2-1)} ?

The first derivative of f(x)= 1/(\sqrt[3]{x^2-1)} is -(2x)/(3(x^2-1)^{4/3)}

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derivative of

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

Solution steps

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{\partial}{\partial x} (y sin (3 x))

\int \frac{- 8 x + 5}{x ^{3} + 4 x} d x

x \to 36 lim (x^{\frac{3}{2}})

y^{^{''}} + x = 7 y

\frac{\partial}{\partial y} (2 x e^{- 2 x y})

What is the derivative of f(x)= 1/(\sqrt[3]{x^2-1)} ?
The derivative of f(x)= 1/(\sqrt[3]{x^2-1)} is -(2x)/(3(x^2-1)^{4/3)}
What is the first derivative of f(x)= 1/(\sqrt[3]{x^2-1)} ?
The first derivative of f(x)= 1/(\sqrt[3]{x^2-1)} is -(2x)/(3(x^2-1)^{4/3)}