d/(dt)(ln(((a+bt)/(a-bt))^{1/2}))

Lösung

\frac{d}{d t} (ln ((\frac{a + b t}{a - b t})^{\frac{1}{2}}))

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

Schritte zur Lösung

\frac{d}{d t} (ln ((\frac{a + b t}{a - b t})^{\frac{1}{2}}))

Wende die Kettenregel an:

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

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

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

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

Vereinfache

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

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

d er i v a t i v e arctan (9 t)

\int_{0}^{2} x^{2} e^{- 9 x} d x

\frac{d}{d x} (6 (sec (x) - tan (x))^{\frac{3}{2}})

a re a x^{2} - 11, 10 x, [- 11, 0] y = 0

\frac{d}{d x} (- 1 0^{3 x^{2} - 4})