Derivatan av $$$\frac{v \left(\ln\left(b\right) + 1\right)}{\ln\left(b\right)}$$$ med avseende på $$$v$$$
Relaterade kalkylatorer: Kalkylator för logaritmisk derivering, Räknare för implicit derivering med steg
Din inmatning
Bestäm $$$\frac{d}{dv} \left(\frac{v \left(\ln\left(b\right) + 1\right)}{\ln\left(b\right)}\right)$$$.
Lösning
Tillämpa konstantfaktorregeln $$$\frac{d}{dv} \left(c f{\left(v \right)}\right) = c \frac{d}{dv} \left(f{\left(v \right)}\right)$$$ med $$$c = \frac{\ln\left(b\right) + 1}{\ln\left(b\right)}$$$ och $$$f{\left(v \right)} = v$$$:
$${\color{red}\left(\frac{d}{dv} \left(\frac{v \left(\ln\left(b\right) + 1\right)}{\ln\left(b\right)}\right)\right)} = {\color{red}\left(\frac{\ln\left(b\right) + 1}{\ln\left(b\right)} \frac{d}{dv} \left(v\right)\right)}$$Tillämpa potensregeln $$$\frac{d}{dv} \left(v^{n}\right) = n v^{n - 1}$$$ med $$$n = 1$$$, det vill säga $$$\frac{d}{dv} \left(v\right) = 1$$$:
$$\frac{\left(\ln\left(b\right) + 1\right) {\color{red}\left(\frac{d}{dv} \left(v\right)\right)}}{\ln\left(b\right)} = \frac{\left(\ln\left(b\right) + 1\right) {\color{red}\left(1\right)}}{\ln\left(b\right)}$$Förenkla:
$$\frac{\ln\left(b\right) + 1}{\ln\left(b\right)} = 1 + \frac{1}{\ln\left(b\right)}$$Alltså, $$$\frac{d}{dv} \left(\frac{v \left(\ln\left(b\right) + 1\right)}{\ln\left(b\right)}\right) = 1 + \frac{1}{\ln\left(b\right)}$$$.
Svar
$$$\frac{d}{dv} \left(\frac{v \left(\ln\left(b\right) + 1\right)}{\ln\left(b\right)}\right) = 1 + \frac{1}{\ln\left(b\right)}$$$A