What is the derivative of 1/(\sqrt[3]{x^2+x+5}) ?

The derivative of 1/(\sqrt[3]{x^2+x+5}) is -(2x+1)/(3(x^2+x+5)^{4/3)}

What is the first derivative of 1/(\sqrt[3]{x^2+x+5}) ?

The first derivative of 1/(\sqrt[3]{x^2+x+5}) is -(2x+1)/(3(x^2+x+5)^{4/3)}

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- \frac{2 x + 1}{3 ( x ^{2} + x + 5 ) ^{\frac{4}{3}}}

Solution steps

Apply exponent rule:

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

= \frac{d}{d x} ((x^{2} + x + 5)^{- \frac{1}{3}})

Apply the chain rule:

- \frac{1}{3 ( x ^{2} + x + 5 ) ^{\frac{4}{3}}} \frac{d}{d x} (x^{2} + x + 5)

= - \frac{1}{3 ( x ^{2} + x + 5 ) ^{\frac{4}{3}}} \frac{d}{d x} (x^{2} + x + 5)

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

= - \frac{1}{3 ( x ^{2} + x + 5 ) ^{\frac{4}{3}}} (2 x + 1)

Simplify

- \frac{1}{3 ( x ^{2} + x + 5 ) ^{\frac{4}{3}}} (2 x + 1) : - \frac{2 x + 1}{3 ( x ^{2} + x + 5 ) ^{\frac{4}{3}}}

= - \frac{2 x + 1}{3 ( x ^{2} + x + 5 ) ^{\frac{4}{3}}}

\int_{0}^{2} e^{x} - e^{- x} d x

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

y = 3 x^{2} + 2 xx a

x \to 0 lim (x^{6} (cos (\frac{9}{x})))

y = e^{- 8 x}, x \geq 0

What is the derivative of 1/(\sqrt[3]{x^2+x+5}) ?
The derivative of 1/(\sqrt[3]{x^2+x+5}) is -(2x+1)/(3(x^2+x+5)^{4/3)}
What is the first derivative of 1/(\sqrt[3]{x^2+x+5}) ?
The first derivative of 1/(\sqrt[3]{x^2+x+5}) is -(2x+1)/(3(x^2+x+5)^{4/3)}