Funktion $$$\ln\left(\frac{t}{t + 1}\right)$$$ derivaatta

Funktio $$$\ln\left(\frac{t}{t + 1}\right)$$$ on kahden funktion $$$f{\left(u \right)} = \ln\left(u\right)$$$ ja $$$g{\left(t \right)} = \frac{t}{t + 1}$$$ yhdistelmä $$$f{\left(g{\left(t \right)} \right)}$$$.

Sovella ketjusääntöä $$$\frac{d}{dt} \left(f{\left(g{\left(t \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dt} \left(g{\left(t \right)}\right)$$$:

$${\color{red}\left(\frac{d}{dt} \left(\ln\left(\frac{t}{t + 1}\right)\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dt} \left(\frac{t}{t + 1}\right)\right)}$$

Luonnollisen logaritmin derivaatta on $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:

$${\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dt} \left(\frac{t}{t + 1}\right) = {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dt} \left(\frac{t}{t + 1}\right)$$

Palaa alkuperäiseen muuttujaan:

$$\frac{\frac{d}{dt} \left(\frac{t}{t + 1}\right)}{{\color{red}\left(u\right)}} = \frac{\frac{d}{dt} \left(\frac{t}{t + 1}\right)}{{\color{red}\left(\frac{t}{t + 1}\right)}}$$

Sovella osamääräsääntöä $$$\frac{d}{dt} \left(\frac{f{\left(t \right)}}{g{\left(t \right)}}\right) = \frac{\frac{d}{dt} \left(f{\left(t \right)}\right) g{\left(t \right)} - f{\left(t \right)} \frac{d}{dt} \left(g{\left(t \right)}\right)}{g^{2}{\left(t \right)}}$$$ funktioille $$$f{\left(t \right)} = t$$$ ja $$$g{\left(t \right)} = t + 1$$$:

$$\frac{\left(t + 1\right) {\color{red}\left(\frac{d}{dt} \left(\frac{t}{t + 1}\right)\right)}}{t} = \frac{\left(t + 1\right) {\color{red}\left(\frac{\frac{d}{dt} \left(t\right) \left(t + 1\right) - t \frac{d}{dt} \left(t + 1\right)}{\left(t + 1\right)^{2}}\right)}}{t}$$

Sovella potenssisääntöä $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$ käyttäen $$$n = 1$$$, toisin sanoen, $$$\frac{d}{dt} \left(t\right) = 1$$$:

$$\frac{- t \frac{d}{dt} \left(t + 1\right) + \left(t + 1\right) {\color{red}\left(\frac{d}{dt} \left(t\right)\right)}}{t \left(t + 1\right)} = \frac{- t \frac{d}{dt} \left(t + 1\right) + \left(t + 1\right) {\color{red}\left(1\right)}}{t \left(t + 1\right)}$$

Summan/erotuksen derivaatta on derivaattojen summa/erotus:

$$\frac{- t {\color{red}\left(\frac{d}{dt} \left(t + 1\right)\right)} + t + 1}{t \left(t + 1\right)} = \frac{- t {\color{red}\left(\frac{d}{dt} \left(t\right) + \frac{d}{dt} \left(1\right)\right)} + t + 1}{t \left(t + 1\right)}$$

Vakion derivaatta on $$$0$$$:

$$\frac{- t \left({\color{red}\left(\frac{d}{dt} \left(1\right)\right)} + \frac{d}{dt} \left(t\right)\right) + t + 1}{t \left(t + 1\right)} = \frac{- t \left({\color{red}\left(0\right)} + \frac{d}{dt} \left(t\right)\right) + t + 1}{t \left(t + 1\right)}$$

Sovella potenssisääntöä $$$\frac{d}{dt} \left(t^{n}\right) = n t^{n - 1}$$$ käyttäen $$$n = 1$$$, toisin sanoen, $$$\frac{d}{dt} \left(t\right) = 1$$$:

$$\frac{- t {\color{red}\left(\frac{d}{dt} \left(t\right)\right)} + t + 1}{t \left(t + 1\right)} = \frac{- t {\color{red}\left(1\right)} + t + 1}{t \left(t + 1\right)}$$

Näin ollen, $$$\frac{d}{dt} \left(\ln\left(\frac{t}{t + 1}\right)\right) = \frac{1}{t \left(t + 1\right)}$$$.