What is the derivative of ln(ax^2+b) ?

The derivative of ln(ax^2+b) is (2ax)/(ax^2+b)

What is the first derivative of ln(ax^2+b) ?

The first derivative of ln(ax^2+b) is (2ax)/(ax^2+b)

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\frac{d}{d x} (ln (a x^{2} + b))

\frac{2 a x}{a x ^{2} + b}

Solution steps

\frac{d}{d x} (ln (a x^{2} + b))

Apply the chain rule:

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

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

\frac{d}{d x} (a x^{2} + b) = 2 a x

= \frac{1}{a x ^{2} + b} \cdot 2 a x

Simplify

\frac{1}{a x ^{2} + b} \cdot 2 a x : \frac{2 a x}{a x ^{2} + b}

= \frac{2 a x}{a x ^{2} + b}

y \to 1 lim (85)

\frac{d}{d y} (2 e^{x y^{2}})

y^{^{'}} = 3 y + x

f (x) = \frac{x}{1 + x}

n = 0 \sum \infty \frac{e ^{n} + 5}{e ^{2 n}}

What is the derivative of ln(ax^2+b) ?
The derivative of ln(ax^2+b) is (2ax)/(ax^2+b)
What is the first derivative of ln(ax^2+b) ?
The first derivative of ln(ax^2+b) is (2ax)/(ax^2+b)