What is the derivative of ln(3x^3+4x^2+x+4) ?

The derivative of ln(3x^3+4x^2+x+4) is (9x^2+8x+1)/(3x^3+4x^2+x+4)

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

The first derivative of ln(3x^3+4x^2+x+4) is (9x^2+8x+1)/(3x^3+4x^2+x+4)

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

\frac{9 x ^{2} + 8 x + 1}{3 x ^{3} + 4 x ^{2} + x + 4}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (3 x^{3} + 4 x^{2} + x + 4) = 9 x^{2} + 8 x + 1

= \frac{1}{3 x ^{3} + 4 x ^{2} + x + 4} (9 x^{2} + 8 x + 1)

Simplify

\frac{1}{3 x ^{3} + 4 x ^{2} + x + 4} (9 x^{2} + 8 x + 1) : \frac{9 x ^{2} + 8 x + 1}{3 x ^{3} + 4 x ^{2} + x + 4}

= \frac{9 x ^{2} + 8 x + 1}{3 x ^{3} + 4 x ^{2} + x + 4}

\frac{d}{d t} (\frac{e ^{2 t}}{1 - t ^{2}})

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

\frac{d}{d x} (cos (\frac{3}{2} x))

n = 1 \sum \infty \frac{1 6 ^{\frac{n}{2}}}{2 ^{2 n}}

\int \frac{1}{e ^{- y}} d y

What is the derivative of ln(3x^3+4x^2+x+4) ?
The derivative of ln(3x^3+4x^2+x+4) is (9x^2+8x+1)/(3x^3+4x^2+x+4)
What is the first derivative of ln(3x^3+4x^2+x+4) ?
The first derivative of ln(3x^3+4x^2+x+4) is (9x^2+8x+1)/(3x^3+4x^2+x+4)