What is the d/(dt)(ln(sqrt((a+bt)/(a-bt)))) ?

The d/(dt)(ln(sqrt((a+bt)/(a-bt)))) is (ab)/((-bt+a)(bt+a))

What is the first d/(dt)(ln(sqrt((a+bt)/(a-bt)))) ?

The first d/(dt)(ln(sqrt((a+bt)/(a-bt)))) is (ab)/((-bt+a)(bt+a))

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\frac{d}{d t} (ln (\frac{a + b t}{a - b t}))

\frac{ab}{( - b t + a ) ( b t + a )}

Solution steps

\frac{d}{d t} (ln (\frac{a + b t}{a - b t}))

Apply the chain rule:

\frac{1}{\frac{a + b t}{a - b t}} \frac{d}{d t} (\frac{a + b t}{a - b t})

= \frac{1}{\frac{a + b t}{a - b t}} \frac{d}{d t} (\frac{a + b t}{a - b t})

\frac{d}{d t} (\frac{a + b t}{a - b t}) = \frac{ab}{b t + a ( - b t + a ) ^{\frac{3}{2}}}

= \frac{1}{\frac{a + b t}{a - b t}} \cdot \frac{ab}{b t + a ( - b t + a ) ^{\frac{3}{2}}}

Simplify

\frac{1}{\frac{a + b t}{a - b t}} \cdot \frac{ab}{b t + a ( - b t + a ) ^{\frac{3}{2}}} : \frac{ab}{( - b t + a ) ( b t + a )}

= \frac{ab}{( - b t + a ) ( b t + a )}

\frac{d}{d s} (\frac{2}{s ^{2} + 4})

\int_{1}^{2} x^{2} d x

\int_{0}^{2} \frac{x}{16 - x ^{2}} d x

\int \frac{5}{2 x + 7} d x

x \to 2 lim (\frac{x + 4}{3 x - 1})

What is the d/(dt)(ln(sqrt((a+bt)/(a-bt)))) ?
The d/(dt)(ln(sqrt((a+bt)/(a-bt)))) is (ab)/((-bt+a)(bt+a))
What is the first d/(dt)(ln(sqrt((a+bt)/(a-bt)))) ?
The first d/(dt)(ln(sqrt((a+bt)/(a-bt)))) is (ab)/((-bt+a)(bt+a))