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

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

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

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

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

Solution steps

Apply the chain rule:

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

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

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

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

Simplify

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

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

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

(1 + x^{2}) (1 + y^{2}) d x - x y d y = 0

\int (6 sin (3 x) - cos (2 x)) d x

\int x - 4 d x

x \to 0 + lim (\frac{3}{x ^{2} - x})

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