What is the derivative of \sqrt[3]{x^3+6x+7} ?

The derivative of \sqrt[3]{x^3+6x+7} is (x^2+2)/((x^3+6x+7)^{2/3)}

What is the first derivative of \sqrt[3]{x^3+6x+7} ?

The first derivative of \sqrt[3]{x^3+6x+7} is (x^2+2)/((x^3+6x+7)^{2/3)}

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\frac{x ^{2} + 2}{( x ^{3} + 6 x + 7 ) ^{\frac{2}{3}}}

Solution steps

Apply the chain rule:

\frac{1}{3 ( x ^{3} + 6 x + 7 ) ^{\frac{2}{3}}} \frac{d}{d x} (x^{3} + 6 x + 7)

= \frac{1}{3 ( x ^{3} + 6 x + 7 ) ^{\frac{2}{3}}} \frac{d}{d x} (x^{3} + 6 x + 7)

\frac{d}{d x} (x^{3} + 6 x + 7) = 3 x^{2} + 6

= \frac{1}{3 ( x ^{3} + 6 x + 7 ) ^{\frac{2}{3}}} (3 x^{2} + 6)

Simplify

\frac{1}{3 ( x ^{3} + 6 x + 7 ) ^{\frac{2}{3}}} (3 x^{2} + 6) : \frac{x ^{2} + 2}{( x ^{3} + 6 x + 7 ) ^{\frac{2}{3}}}

= \frac{x ^{2} + 2}{( x ^{3} + 6 x + 7 ) ^{\frac{2}{3}}}

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

\int \frac{x ^{3} + x ^{2} + x + 2}{x ^{4} + 3 x ^{2} + 2} d x

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

\frac{5}{( 2 x ) ^{3}}

\int_{0}^{\frac{1}{4}} x arcsin (4 x) d x

What is the derivative of \sqrt[3]{x^3+6x+7} ?
The derivative of \sqrt[3]{x^3+6x+7} is (x^2+2)/((x^3+6x+7)^{2/3)}
What is the first derivative of \sqrt[3]{x^3+6x+7} ?
The first derivative of \sqrt[3]{x^3+6x+7} is (x^2+2)/((x^3+6x+7)^{2/3)}