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

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

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

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

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- \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{2}{s ^{2}} \cdot e^{- s}

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

\frac{d}{d x} (x^{2} + 6 x + 3)

\frac{d}{d x} (3 x^{2} - 12 x + 9)

11 t^{2} y^{^{''}} - 22 y = 9 t^{2} - 9

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