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

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

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

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

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\int e^{\frac{- x + 2}{3}} d x

y^{^{''}} - 16 y = \frac{16 x}{e ^{4 x}}

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

\int e^{- \frac{1}{2} x^{2}} d x

y^{^{''}} + 2 y + y = 2 e^{- t}

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