What is the derivative of ln(x^2-9) ?

The derivative of ln(x^2-9) is (2x)/(x^2-9)

What is the first derivative of ln(x^2-9) ?

The first derivative of ln(x^2-9) is (2x)/(x^2-9)

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\frac{d}{d x} (ln (x^{2} - 9))

\frac{2 x}{x ^{2} - 9}

Solution steps

\frac{d}{d x} (ln (x^{2} - 9))

Apply the chain rule:

\frac{1}{x ^{2} - 9} \frac{d}{d x} (x^{2} - 9)

= \frac{1}{x ^{2} - 9} \frac{d}{d x} (x^{2} - 9)

\frac{d}{d x} (x^{2} - 9) = 2 x

= \frac{1}{x ^{2} - 9} \cdot 2 x

Simplify

\frac{1}{x ^{2} - 9} \cdot 2 x : \frac{2 x}{x ^{2} - 9}

= \frac{2 x}{x ^{2} - 9}

\int 6 x ln (3 x) d x

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

\int x (x + 2) d x

\int \frac{e ^{x} e ^{x} - 1}{e ^{x} + 3} d x

\frac{d y}{d x} = \frac{1}{x ^{2} + 1} - \frac{1}{1 - x ^{2}}

What is the derivative of ln(x^2-9) ?
The derivative of ln(x^2-9) is (2x)/(x^2-9)
What is the first derivative of ln(x^2-9) ?
The first derivative of ln(x^2-9) is (2x)/(x^2-9)