What is the derivative of ln(3x^3+6x^2+2x+6) ?

The derivative of ln(3x^3+6x^2+2x+6) is (9x^2+12x+2)/(3x^3+6x^2+2x+6)

What is the first derivative of ln(3x^3+6x^2+2x+6) ?

The first derivative of ln(3x^3+6x^2+2x+6) is (9x^2+12x+2)/(3x^3+6x^2+2x+6)

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derivative of

ln (3 x^{3} + 6 x^{2} + 2 x + 6)

\frac{9 x ^{2} + 12 x + 2}{3 x ^{3} + 6 x ^{2} + 2 x + 6}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (3 x^{3} + 6 x^{2} + 2 x + 6) = 9 x^{2} + 12 x + 2

= \frac{1}{3 x ^{3} + 6 x ^{2} + 2 x + 6} (9 x^{2} + 12 x + 2)

Simplify

\frac{1}{3 x ^{3} + 6 x ^{2} + 2 x + 6} (9 x^{2} + 12 x + 2) : \frac{9 x ^{2} + 12 x + 2}{3 x ^{3} + 6 x ^{2} + 2 x + 6}

= \frac{9 x ^{2} + 12 x + 2}{3 x ^{3} + 6 x ^{2} + 2 x + 6}

x y^{2} y^{^{'}} = x + 9

\int x^{n} a x^{n + 1} + b d x

r x + d

\frac{\partial}{\partial x} (4 y^{3} sin (5 x))

\frac{s ^{2} - s + 1}{( s + 1 ) ^{2}}

What is the derivative of ln(3x^3+6x^2+2x+6) ?
The derivative of ln(3x^3+6x^2+2x+6) is (9x^2+12x+2)/(3x^3+6x^2+2x+6)
What is the first derivative of ln(3x^3+6x^2+2x+6) ?
The first derivative of ln(3x^3+6x^2+2x+6) is (9x^2+12x+2)/(3x^3+6x^2+2x+6)