What is the d/(ds)(ln(s^2+9)) ?

The d/(ds)(ln(s^2+9)) is (2s)/(s^2+9)

What is the first d/(ds)(ln(s^2+9)) ?

The first d/(ds)(ln(s^2+9)) is (2s)/(s^2+9)

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\frac{d}{d s} (ln (s^{2} + 9))

\frac{2 s}{s ^{2} + 9}

Solution steps

\frac{d}{d s} (ln (s^{2} + 9))

Apply the chain rule:

\frac{1}{s ^{2} + 9} \frac{d}{d s} (s^{2} + 9)

= \frac{1}{s ^{2} + 9} \frac{d}{d s} (s^{2} + 9)

\frac{d}{d s} (s^{2} + 9) = 2 s

= \frac{1}{s ^{2} + 9} \cdot 2 s

Simplify

\frac{1}{s ^{2} + 9} \cdot 2 s : \frac{2 s}{s ^{2} + 9}

= \frac{2 s}{s ^{2} + 9}

4 t^{- \frac{1}{8}}

r (x) = \frac{3}{x ^{4} + 2}

f (x) = \frac{6}{x ^{3}}

\frac{d}{d x} (sin (e^{3 x}))

What is the d/(ds)(ln(s^2+9)) ?
The d/(ds)(ln(s^2+9)) is (2s)/(s^2+9)
What is the first d/(ds)(ln(s^2+9)) ?
The first d/(ds)(ln(s^2+9)) is (2s)/(s^2+9)