Derivatan av $$$\frac{\left(x + 1\right)^{6}}{\left(x^{2} + 8\right)^{6}}$$$

Låt $$$H{\left(x \right)} = \frac{\left(x + 1\right)^{6}}{\left(x^{2} + 8\right)^{6}}$$$.

Ta logaritmen av båda sidorna: $$$\ln\left(H{\left(x \right)}\right) = \ln\left(\frac{\left(x + 1\right)^{6}}{\left(x^{2} + 8\right)^{6}}\right)$$$.

Skriv om högerledet med hjälp av logaritmlagarna: $$$\ln\left(H{\left(x \right)}\right) = 6 \ln\left(x + 1\right) - 6 \ln\left(x^{2} + 8\right)$$$.

Derivera båda leden i ekvationen var för sig: $$$\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right) = \frac{d}{dx} \left(6 \ln\left(x + 1\right) - 6 \ln\left(x^{2} + 8\right)\right)$$$.

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:

$${\color{red}\left(\frac{d}{dx} \left(6 \ln\left(x + 1\right) - 6 \ln\left(x^{2} + 8\right)\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(6 \ln\left(x + 1\right)\right) - \frac{d}{dx} \left(6 \ln\left(x^{2} + 8\right)\right)\right)}$$

Tillämpa konstantfaktorregeln $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ med $$$c = 6$$$ och $$$f{\left(x \right)} = \ln\left(x^{2} + 8\right)$$$:

$$- {\color{red}\left(\frac{d}{dx} \left(6 \ln\left(x^{2} + 8\right)\right)\right)} + \frac{d}{dx} \left(6 \ln\left(x + 1\right)\right) = - {\color{red}\left(6 \frac{d}{dx} \left(\ln\left(x^{2} + 8\right)\right)\right)} + \frac{d}{dx} \left(6 \ln\left(x + 1\right)\right)$$

Tillämpa konstantfaktorregeln $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ med $$$c = 6$$$ och $$$f{\left(x \right)} = \ln\left(x + 1\right)$$$:

$${\color{red}\left(\frac{d}{dx} \left(6 \ln\left(x + 1\right)\right)\right)} - 6 \frac{d}{dx} \left(\ln\left(x^{2} + 8\right)\right) = {\color{red}\left(6 \frac{d}{dx} \left(\ln\left(x + 1\right)\right)\right)} - 6 \frac{d}{dx} \left(\ln\left(x^{2} + 8\right)\right)$$

Funktionen $$$\ln\left(x + 1\right)$$$ är sammansättningen $$$f{\left(g{\left(x \right)} \right)}$$$ av två funktioner $$$f{\left(u \right)} = \ln\left(u\right)$$$ och $$$g{\left(x \right)} = x + 1$$$.

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

$$6 {\color{red}\left(\frac{d}{dx} \left(\ln\left(x + 1\right)\right)\right)} - 6 \frac{d}{dx} \left(\ln\left(x^{2} + 8\right)\right) = 6 {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(x + 1\right)\right)} - 6 \frac{d}{dx} \left(\ln\left(x^{2} + 8\right)\right)$$

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

$$6 {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(x + 1\right) - 6 \frac{d}{dx} \left(\ln\left(x^{2} + 8\right)\right) = 6 {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(x + 1\right) - 6 \frac{d}{dx} \left(\ln\left(x^{2} + 8\right)\right)$$

Återgå till den ursprungliga variabeln:

$$- 6 \frac{d}{dx} \left(\ln\left(x^{2} + 8\right)\right) + \frac{6 \frac{d}{dx} \left(x + 1\right)}{{\color{red}\left(u\right)}} = - 6 \frac{d}{dx} \left(\ln\left(x^{2} + 8\right)\right) + \frac{6 \frac{d}{dx} \left(x + 1\right)}{{\color{red}\left(x + 1\right)}}$$

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

$$- 6 \frac{d}{dx} \left(\ln\left(x^{2} + 8\right)\right) + \frac{6 {\color{red}\left(\frac{d}{dx} \left(x + 1\right)\right)}}{x + 1} = - 6 \frac{d}{dx} \left(\ln\left(x^{2} + 8\right)\right) + \frac{6 {\color{red}\left(\frac{d}{dx} \left(x\right) + \frac{d}{dx} \left(1\right)\right)}}{x + 1}$$

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

$$- 6 \frac{d}{dx} \left(\ln\left(x^{2} + 8\right)\right) + \frac{6 \left({\color{red}\left(\frac{d}{dx} \left(x\right)\right)} + \frac{d}{dx} \left(1\right)\right)}{x + 1} = - 6 \frac{d}{dx} \left(\ln\left(x^{2} + 8\right)\right) + \frac{6 \left({\color{red}\left(1\right)} + \frac{d}{dx} \left(1\right)\right)}{x + 1}$$

Derivatan av en konstant är $$$0$$$:

$$- 6 \frac{d}{dx} \left(\ln\left(x^{2} + 8\right)\right) + \frac{6 \left({\color{red}\left(\frac{d}{dx} \left(1\right)\right)} + 1\right)}{x + 1} = - 6 \frac{d}{dx} \left(\ln\left(x^{2} + 8\right)\right) + \frac{6 \left({\color{red}\left(0\right)} + 1\right)}{x + 1}$$

Funktionen $$$\ln\left(x^{2} + 8\right)$$$ är sammansättningen $$$f{\left(g{\left(x \right)} \right)}$$$ av två funktioner $$$f{\left(u \right)} = \ln\left(u\right)$$$ och $$$g{\left(x \right)} = x^{2} + 8$$$.

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

$$- 6 {\color{red}\left(\frac{d}{dx} \left(\ln\left(x^{2} + 8\right)\right)\right)} + \frac{6}{x + 1} = - 6 {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(x^{2} + 8\right)\right)} + \frac{6}{x + 1}$$

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

$$- 6 {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(x^{2} + 8\right) + \frac{6}{x + 1} = - 6 {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(x^{2} + 8\right) + \frac{6}{x + 1}$$

Återgå till den ursprungliga variabeln:

$$- \frac{6 \frac{d}{dx} \left(x^{2} + 8\right)}{{\color{red}\left(u\right)}} + \frac{6}{x + 1} = - \frac{6 \frac{d}{dx} \left(x^{2} + 8\right)}{{\color{red}\left(x^{2} + 8\right)}} + \frac{6}{x + 1}$$

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

$$- \frac{6 {\color{red}\left(\frac{d}{dx} \left(x^{2} + 8\right)\right)}}{x^{2} + 8} + \frac{6}{x + 1} = - \frac{6 {\color{red}\left(\frac{d}{dx} \left(x^{2}\right) + \frac{d}{dx} \left(8\right)\right)}}{x^{2} + 8} + \frac{6}{x + 1}$$

Derivatan av en konstant är $$$0$$$:

$$- \frac{6 \left({\color{red}\left(\frac{d}{dx} \left(8\right)\right)} + \frac{d}{dx} \left(x^{2}\right)\right)}{x^{2} + 8} + \frac{6}{x + 1} = - \frac{6 \left({\color{red}\left(0\right)} + \frac{d}{dx} \left(x^{2}\right)\right)}{x^{2} + 8} + \frac{6}{x + 1}$$

Tillämpa potensregeln $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ med $$$n = 2$$$:

$$- \frac{6 {\color{red}\left(\frac{d}{dx} \left(x^{2}\right)\right)}}{x^{2} + 8} + \frac{6}{x + 1} = - \frac{6 {\color{red}\left(2 x\right)}}{x^{2} + 8} + \frac{6}{x + 1}$$

Alltså, $$$\frac{d}{dx} \left(6 \ln\left(x + 1\right) - 6 \ln\left(x^{2} + 8\right)\right) = - \frac{12 x}{x^{2} + 8} + \frac{6}{x + 1}$$$.

Således, $$$\frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{H{\left(x \right)}} = - \frac{12 x}{x^{2} + 8} + \frac{6}{x + 1}$$$.

Således, $$$\frac{d}{dx} \left(H{\left(x \right)}\right) = \left(- \frac{12 x}{x^{2} + 8} + \frac{6}{x + 1}\right) H{\left(x \right)} = - \frac{6 \left(x - 2\right) \left(x + 1\right)^{5} \left(x + 4\right)}{\left(x^{2} + 8\right)^{7}}.$$$