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

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

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

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

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

x \to 1 - lim (- \frac{2}{1 - x})

x \to 0 lim (ln (sin (x)))

x, y = sin (x y)

(ln (2 x))^{2}

\int - e^{- 3 x} (x - 3) (x - 6) d x

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