What is the first derivative of ln|t| ?

The first derivative of ln|t| is 1/t

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derivative of ln|t|

derivative of

ln ∣ t ∣

\frac{1}{t}

Solution steps

\frac{d}{d t} (ln ∣ t ∣)

Apply the chain rule:

\frac{1}{∣ t ∣} \frac{d}{d t} (∣ t ∣)

= \frac{1}{∣ t ∣} \frac{d}{d t} (∣ t ∣)

\frac{d}{d t} (∣ t ∣) = \frac{t}{∣ t ∣}

= \frac{1}{∣ t ∣} \cdot \frac{t}{∣ t ∣}

Simplify

\frac{1}{∣ t ∣} \cdot \frac{t}{∣ t ∣} : \frac{1}{t}

= \frac{1}{t}

d er i v a t i v e \frac{( x + 1 )}{x - 1}

t an g e n t y = sin (2 x) ln (x^{2}), (1, 0)

\int sinh (x) cosh (x) d x

t a y l or 1 - x, 0

y^{^{''}} + 3 y^{^{'}} + 2 y = cos (2 x)