What is the derivative of y=ln(x^3+1) ?

The derivative of y=ln(x^3+1) is (3x^2)/(x^3+1)

What is the first derivative of y=ln(x^3+1) ?

The first derivative of y=ln(x^3+1) is (3x^2)/(x^3+1)

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derivative of

y = ln (x^{3} + 1)

\frac{3 x ^{2}}{x ^{3} + 1}

Solution steps

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

Apply the chain rule:

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

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

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

= \frac{1}{x ^{3} + 1} \cdot 3 x^{2}

Simplify

\frac{1}{x ^{3} + 1} \cdot 3 x^{2} : \frac{3 x ^{2}}{x ^{3} + 1}

= \frac{3 x ^{2}}{x ^{3} + 1}

n = 0 \sum \infty \frac{1}{3}

\int_{0}^{1} (8 x^{2} - 2) e^{x} d x

\frac{d}{d t} ((arccos (t))^{2})

\int 2 cos^{3} (4 x) d x

\frac{\partial}{\partial v} (2 uv)

What is the derivative of y=ln(x^3+1) ?
The derivative of y=ln(x^3+1) is (3x^2)/(x^3+1)
What is the first derivative of y=ln(x^3+1) ?
The first derivative of y=ln(x^3+1) is (3x^2)/(x^3+1)