What is the derivative of ln(3x^2+7) ?

The derivative of ln(3x^2+7) is (6x)/(3x^2+7)

What is the first derivative of ln(3x^2+7) ?

The first derivative of ln(3x^2+7) is (6x)/(3x^2+7)

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\frac{d}{d x} (ln (3 x^{2} + 7))

\frac{6 x}{3 x ^{2} + 7}

Solution steps

\frac{d}{d x} (ln (3 x^{2} + 7))

Apply the chain rule:

\frac{1}{3 x ^{2} + 7} \frac{d}{d x} (3 x^{2} + 7)

= \frac{1}{3 x ^{2} + 7} \frac{d}{d x} (3 x^{2} + 7)

\frac{d}{d x} (3 x^{2} + 7) = 6 x

= \frac{1}{3 x ^{2} + 7} \cdot 6 x

Simplify

\frac{1}{3 x ^{2} + 7} \cdot 6 x : \frac{6 x}{3 x ^{2} + 7}

= \frac{6 x}{3 x ^{2} + 7}

f (x) = x - \frac{1}{x}, at x = 1

y^{^{''}} + 81 y = 4 tan (9 x)

x \to 4 lim (\frac{x}{4} + 1)

\int 6.5 x d x

\frac{( 10 s + 3 )}{s ^{2} + 4 s + 29}

What is the derivative of ln(3x^2+7) ?
The derivative of ln(3x^2+7) is (6x)/(3x^2+7)
What is the first derivative of ln(3x^2+7) ?
The first derivative of ln(3x^2+7) is (6x)/(3x^2+7)