Derivatan av $$$\ln\left(x\right) \sin{\left(9 x \right)}$$$

Låt $$$H{\left(x \right)} = \ln\left(x\right) \sin{\left(9 x \right)}$$$.

Ta logaritmen av båda sidorna: $$$\ln\left(H{\left(x \right)}\right) = \ln\left(\ln\left(x\right) \sin{\left(9 x \right)}\right)$$$.

Skriv om högerledet med hjälp av logaritmlagarna: $$$\ln\left(H{\left(x \right)}\right) = \ln\left(\ln\left(x\right)\right) + \ln\left(\sin{\left(9 x \right)}\right)$$$.

Derivera båda leden i ekvationen var för sig: $$$\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right) = \frac{d}{dx} \left(\ln\left(\ln\left(x\right)\right) + \ln\left(\sin{\left(9 x \right)}\right)\right)$$$.

Derivera ekvationens vänsterled.

Funktionen $$$\ln\left(H{\left(x \right)}\right)$$$ är sammansättningen $$$f{\left(g{\left(x \right)} \right)}$$$ av två funktioner $$$f{\left(u \right)} = \ln\left(u\right)$$$ och $$$g{\left(x \right)} = H{\left(x \right)}$$$.

Tillämpa kedjeregeln $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

$${\color{red}\left(\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(H{\left(x \right)}\right)\right)}$$

Derivatan av den naturliga logaritmen är $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:

$${\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(H{\left(x \right)}\right) = {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(H{\left(x \right)}\right)$$

Återgå till den ursprungliga variabeln:

$$\frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{{\color{red}\left(u\right)}} = \frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{{\color{red}\left(H{\left(x \right)}\right)}}$$

Alltså, $$$\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right) = \frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{H{\left(x \right)}}$$$.

Derivera ekvationens högerled.

Derivatan av en summa/differens är summan/differensen av derivatorna:

$${\color{red}\left(\frac{d}{dx} \left(\ln\left(\ln\left(x\right)\right) + \ln\left(\sin{\left(9 x \right)}\right)\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(\ln\left(\ln\left(x\right)\right)\right) + \frac{d}{dx} \left(\ln\left(\sin{\left(9 x \right)}\right)\right)\right)}$$

Funktionen $$$\ln\left(\ln\left(x\right)\right)$$$ är sammansättningen $$$f{\left(g{\left(x \right)} \right)}$$$ av två funktioner $$$f{\left(u \right)} = \ln\left(u\right)$$$ och $$$g{\left(x \right)} = \ln\left(x\right)$$$.

Tillämpa kedjeregeln $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

$${\color{red}\left(\frac{d}{dx} \left(\ln\left(\ln\left(x\right)\right)\right)\right)} + \frac{d}{dx} \left(\ln\left(\sin{\left(9 x \right)}\right)\right) = {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(\ln\left(x\right)\right)\right)} + \frac{d}{dx} \left(\ln\left(\sin{\left(9 x \right)}\right)\right)$$

Derivatan av den naturliga logaritmen är $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:

$${\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(\ln\left(x\right)\right) + \frac{d}{dx} \left(\ln\left(\sin{\left(9 x \right)}\right)\right) = {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(\ln\left(x\right)\right) + \frac{d}{dx} \left(\ln\left(\sin{\left(9 x \right)}\right)\right)$$

Återgå till den ursprungliga variabeln:

$$\frac{d}{dx} \left(\ln\left(\sin{\left(9 x \right)}\right)\right) + \frac{\frac{d}{dx} \left(\ln\left(x\right)\right)}{{\color{red}\left(u\right)}} = \frac{d}{dx} \left(\ln\left(\sin{\left(9 x \right)}\right)\right) + \frac{\frac{d}{dx} \left(\ln\left(x\right)\right)}{{\color{red}\left(\ln\left(x\right)\right)}}$$

Funktionen $$$\ln\left(\sin{\left(9 x \right)}\right)$$$ är sammansättningen $$$f{\left(g{\left(x \right)} \right)}$$$ av två funktioner $$$f{\left(u \right)} = \ln\left(u\right)$$$ och $$$g{\left(x \right)} = \sin{\left(9 x \right)}$$$.

Tillämpa kedjeregeln $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

$${\color{red}\left(\frac{d}{dx} \left(\ln\left(\sin{\left(9 x \right)}\right)\right)\right)} + \frac{\frac{d}{dx} \left(\ln\left(x\right)\right)}{\ln\left(x\right)} = {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(\sin{\left(9 x \right)}\right)\right)} + \frac{\frac{d}{dx} \left(\ln\left(x\right)\right)}{\ln\left(x\right)}$$

Derivatan av den naturliga logaritmen är $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:

$${\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(\sin{\left(9 x \right)}\right) + \frac{\frac{d}{dx} \left(\ln\left(x\right)\right)}{\ln\left(x\right)} = {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(\sin{\left(9 x \right)}\right) + \frac{\frac{d}{dx} \left(\ln\left(x\right)\right)}{\ln\left(x\right)}$$

Återgå till den ursprungliga variabeln:

$$\frac{\frac{d}{dx} \left(\sin{\left(9 x \right)}\right)}{{\color{red}\left(u\right)}} + \frac{\frac{d}{dx} \left(\ln\left(x\right)\right)}{\ln\left(x\right)} = \frac{\frac{d}{dx} \left(\sin{\left(9 x \right)}\right)}{{\color{red}\left(\sin{\left(9 x \right)}\right)}} + \frac{\frac{d}{dx} \left(\ln\left(x\right)\right)}{\ln\left(x\right)}$$

Funktionen $$$\sin{\left(9 x \right)}$$$ är sammansättningen $$$f{\left(g{\left(x \right)} \right)}$$$ av två funktioner $$$f{\left(u \right)} = \sin{\left(u \right)}$$$ och $$$g{\left(x \right)} = 9 x$$$.

Tillämpa kedjeregeln $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

$$\frac{{\color{red}\left(\frac{d}{dx} \left(\sin{\left(9 x \right)}\right)\right)}}{\sin{\left(9 x \right)}} + \frac{\frac{d}{dx} \left(\ln\left(x\right)\right)}{\ln\left(x\right)} = \frac{{\color{red}\left(\frac{d}{du} \left(\sin{\left(u \right)}\right) \frac{d}{dx} \left(9 x\right)\right)}}{\sin{\left(9 x \right)}} + \frac{\frac{d}{dx} \left(\ln\left(x\right)\right)}{\ln\left(x\right)}$$

Derivatan av sinus är $$$\frac{d}{du} \left(\sin{\left(u \right)}\right) = \cos{\left(u \right)}$$$:

$$\frac{{\color{red}\left(\frac{d}{du} \left(\sin{\left(u \right)}\right)\right)} \frac{d}{dx} \left(9 x\right)}{\sin{\left(9 x \right)}} + \frac{\frac{d}{dx} \left(\ln\left(x\right)\right)}{\ln\left(x\right)} = \frac{{\color{red}\left(\cos{\left(u \right)}\right)} \frac{d}{dx} \left(9 x\right)}{\sin{\left(9 x \right)}} + \frac{\frac{d}{dx} \left(\ln\left(x\right)\right)}{\ln\left(x\right)}$$

Återgå till den ursprungliga variabeln:

$$\frac{\cos{\left({\color{red}\left(u\right)} \right)} \frac{d}{dx} \left(9 x\right)}{\sin{\left(9 x \right)}} + \frac{\frac{d}{dx} \left(\ln\left(x\right)\right)}{\ln\left(x\right)} = \frac{\cos{\left({\color{red}\left(9 x\right)} \right)} \frac{d}{dx} \left(9 x\right)}{\sin{\left(9 x \right)}} + \frac{\frac{d}{dx} \left(\ln\left(x\right)\right)}{\ln\left(x\right)}$$

Tillämpa konstantfaktorregeln $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ med $$$c = 9$$$ och $$$f{\left(x \right)} = x$$$:

$$\frac{\cos{\left(9 x \right)} {\color{red}\left(\frac{d}{dx} \left(9 x\right)\right)}}{\sin{\left(9 x \right)}} + \frac{\frac{d}{dx} \left(\ln\left(x\right)\right)}{\ln\left(x\right)} = \frac{\cos{\left(9 x \right)} {\color{red}\left(9 \frac{d}{dx} \left(x\right)\right)}}{\sin{\left(9 x \right)}} + \frac{\frac{d}{dx} \left(\ln\left(x\right)\right)}{\ln\left(x\right)}$$

Derivatan av den naturliga logaritmen är $$$\frac{d}{dx} \left(\ln\left(x\right)\right) = \frac{1}{x}$$$:

$$\frac{9 \cos{\left(9 x \right)} \frac{d}{dx} \left(x\right)}{\sin{\left(9 x \right)}} + \frac{{\color{red}\left(\frac{d}{dx} \left(\ln\left(x\right)\right)\right)}}{\ln\left(x\right)} = \frac{9 \cos{\left(9 x \right)} \frac{d}{dx} \left(x\right)}{\sin{\left(9 x \right)}} + \frac{{\color{red}\left(\frac{1}{x}\right)}}{\ln\left(x\right)}$$

Tillämpa potensregeln $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ med $$$n = 1$$$, det vill säga $$$\frac{d}{dx} \left(x\right) = 1$$$:

$$\frac{9 \cos{\left(9 x \right)} {\color{red}\left(\frac{d}{dx} \left(x\right)\right)}}{\sin{\left(9 x \right)}} + \frac{1}{x \ln\left(x\right)} = \frac{9 \cos{\left(9 x \right)} {\color{red}\left(1\right)}}{\sin{\left(9 x \right)}} + \frac{1}{x \ln\left(x\right)}$$

Förenkla:

$$\frac{9 \cos{\left(9 x \right)}}{\sin{\left(9 x \right)}} + \frac{1}{x \ln\left(x\right)} = \frac{9}{\tan{\left(9 x \right)}} + \frac{1}{x \ln\left(x\right)}$$

Alltså, $$$\frac{d}{dx} \left(\ln\left(\ln\left(x\right)\right) + \ln\left(\sin{\left(9 x \right)}\right)\right) = \frac{9}{\tan{\left(9 x \right)}} + \frac{1}{x \ln\left(x\right)}$$$.

Således, $$$\frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{H{\left(x \right)}} = \frac{9}{\tan{\left(9 x \right)}} + \frac{1}{x \ln\left(x\right)}$$$.

Således, $$$\frac{d}{dx} \left(H{\left(x \right)}\right) = \left(\frac{9}{\tan{\left(9 x \right)}} + \frac{1}{x \ln\left(x\right)}\right) H{\left(x \right)} = 9 \ln\left(x\right) \cos{\left(9 x \right)} + \frac{\sin{\left(9 x \right)}}{x}$$$.