What is the derivative of ln(2x^3+8x) ?

The derivative of ln(2x^3+8x) is (3x^2+4)/(x(x^2+4))

What is the first derivative of ln(2x^3+8x) ?

The first derivative of ln(2x^3+8x) is (3x^2+4)/(x(x^2+4))

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\frac{d}{d x} (ln (2 x^{3} + 8 x))

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

Solution steps

\frac{d}{d x} (ln (2 x^{3} + 8 x))

Apply the chain rule:

\frac{1}{2 x ^{3} + 8 x} \frac{d}{d x} (2 x^{3} + 8 x)

= \frac{1}{2 x ^{3} + 8 x} \frac{d}{d x} (2 x^{3} + 8 x)

\frac{d}{d x} (2 x^{3} + 8 x) = 6 x^{2} + 8

= \frac{1}{2 x ^{3} + 8 x} (6 x^{2} + 8)

Simplify

\frac{1}{2 x ^{3} + 8 x} (6 x^{2} + 8) : \frac{3 x ^{2} + 4}{x ( x ^{2} + 4 )}

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

\int 8^{2 x + 6} d x

\frac{8}{0.65 s + 1}

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

\int \frac{3 x - 4}{x ^{2} - x - 6} d x

\frac{d}{d x} (6 x - 3 x)

What is the derivative of ln(2x^3+8x) ?
The derivative of ln(2x^3+8x) is (3x^2+4)/(x(x^2+4))
What is the first derivative of ln(2x^3+8x) ?
The first derivative of ln(2x^3+8x) is (3x^2+4)/(x(x^2+4))