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

The derivative of ln(6x^3+3x^2+3x+3) is (6x^2+2x+1)/(2x^3+x^2+x+1)

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

The first derivative of ln(6x^3+3x^2+3x+3) is (6x^2+2x+1)/(2x^3+x^2+x+1)

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derivative of

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

\frac{6 x ^{2} + 2 x + 1}{2 x ^{3} + x ^{2} + x + 1}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (6 x^{3} + 3 x^{2} + 3 x + 3) = 18 x^{2} + 6 x + 3

= \frac{1}{6 x ^{3} + 3 x ^{2} + 3 x + 3} (18 x^{2} + 6 x + 3)

Simplify

\frac{1}{6 x ^{3} + 3 x ^{2} + 3 x + 3} (18 x^{2} + 6 x + 3) : \frac{6 x ^{2} + 2 x + 1}{2 x ^{3} + x ^{2} + x + 1}

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

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

f (x) = 9 + 5 x + 2 x e^{x}, at x = 0

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

f (x) = 19 - x^{2}, at x = 3

x y^{^{'}} - 2 y^{^{'}} - (x - 2) sin (x) + y = 0

What is the derivative of ln(6x^3+3x^2+3x+3) ?
The derivative of ln(6x^3+3x^2+3x+3) is (6x^2+2x+1)/(2x^3+x^2+x+1)
What is the first derivative of ln(6x^3+3x^2+3x+3) ?
The first derivative of ln(6x^3+3x^2+3x+3) is (6x^2+2x+1)/(2x^3+x^2+x+1)