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

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

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

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

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

Solution steps

Apply exponent rule:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d y}{d x} = \frac{x ^{4} y ^{4} + 4 x ^{4}}{y ^{3}}

\int \frac{1}{x ^{3} 4 x ^{2} - 9} d x

\int csc^{2} (6 θ) cot (6 θ) d θ

x \to \infty lim (x \cdot x)

x \to 0 lim (\frac{sin ( 15 x )}{x})

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