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

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

Neem de logaritme van beide zijden: $$$\ln\left(H{\left(x \right)}\right) = \ln\left(\frac{\left(x + 1\right)^{6}}{\left(x^{2} + 8\right)^{6}}\right)$$$.

Herschrijf het rechterlid met behulp van de eigenschappen van logaritmen: $$$\ln\left(H{\left(x \right)}\right) = 6 \ln\left(x + 1\right) - 6 \ln\left(x^{2} + 8\right)$$$.

Differentieer afzonderlijk beide zijden van de vergelijking: $$$\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right) = \frac{d}{dx} \left(6 \ln\left(x + 1\right) - 6 \ln\left(x^{2} + 8\right)\right)$$$.

Differentieer het linkerlid van de vergelijking.

De functie $$$\ln\left(H{\left(x \right)}\right)$$$ is de samenstelling $$$f{\left(g{\left(x \right)} \right)}$$$ van twee functies $$$f{\left(u \right)} = \ln\left(u\right)$$$ en $$$g{\left(x \right)} = H{\left(x \right)}$$$.

Pas de kettingregel $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ toe:

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

De afgeleide van de natuurlijke logaritme is $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:

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

Keer terug naar de oorspronkelijke variabele:

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

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

Differentieer het rechterlid van de vergelijking.

De afgeleide van een som/verschil is de som/het verschil van de afgeleiden:

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

Pas de regel van de constante factor $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ toe met $$$c = 6$$$ en $$$f{\left(x \right)} = \ln\left(x^{2} + 8\right)$$$:

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

De functie $$$\ln\left(x^{2} + 8\right)$$$ is de samenstelling $$$f{\left(g{\left(x \right)} \right)}$$$ van twee functies $$$f{\left(u \right)} = \ln\left(u\right)$$$ en $$$g{\left(x \right)} = x^{2} + 8$$$.

Pas de kettingregel $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ toe:

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

De afgeleide van de natuurlijke logaritme is $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:

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

Keer terug naar de oorspronkelijke variabele:

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

De afgeleide van een som/verschil is de som/het verschil van de afgeleiden:

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

De afgeleide van een constante is $$$0$$$:

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

Pas de machtsregel $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ toe met $$$n = 2$$$:

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

Pas de regel van de constante factor $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ toe met $$$c = 6$$$ en $$$f{\left(x \right)} = \ln\left(x + 1\right)$$$:

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

De functie $$$\ln\left(x + 1\right)$$$ is de samenstelling $$$f{\left(g{\left(x \right)} \right)}$$$ van twee functies $$$f{\left(u \right)} = \ln\left(u\right)$$$ en $$$g{\left(x \right)} = x + 1$$$.

Pas de kettingregel $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$ toe:

$$- \frac{12 x}{x^{2} + 8} + 6 {\color{red}\left(\frac{d}{dx} \left(\ln\left(x + 1\right)\right)\right)} = - \frac{12 x}{x^{2} + 8} + 6 {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(x + 1\right)\right)}$$

De afgeleide van de natuurlijke logaritme is $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:

$$- \frac{12 x}{x^{2} + 8} + 6 {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(x + 1\right) = - \frac{12 x}{x^{2} + 8} + 6 {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(x + 1\right)$$

Keer terug naar de oorspronkelijke variabele:

$$- \frac{12 x}{x^{2} + 8} + \frac{6 \frac{d}{dx} \left(x + 1\right)}{{\color{red}\left(u\right)}} = - \frac{12 x}{x^{2} + 8} + \frac{6 \frac{d}{dx} \left(x + 1\right)}{{\color{red}\left(x + 1\right)}}$$

De afgeleide van een som/verschil is de som/het verschil van de afgeleiden:

$$- \frac{12 x}{x^{2} + 8} + \frac{6 {\color{red}\left(\frac{d}{dx} \left(x + 1\right)\right)}}{x + 1} = - \frac{12 x}{x^{2} + 8} + \frac{6 {\color{red}\left(\frac{d}{dx} \left(x\right) + \frac{d}{dx} \left(1\right)\right)}}{x + 1}$$

De afgeleide van een constante is $$$0$$$:

$$- \frac{12 x}{x^{2} + 8} + \frac{6 \left({\color{red}\left(\frac{d}{dx} \left(1\right)\right)} + \frac{d}{dx} \left(x\right)\right)}{x + 1} = - \frac{12 x}{x^{2} + 8} + \frac{6 \left({\color{red}\left(0\right)} + \frac{d}{dx} \left(x\right)\right)}{x + 1}$$

Pas de machtsregel $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ toe met $$$n = 1$$$, met andere woorden, $$$\frac{d}{dx} \left(x\right) = 1$$$:

$$- \frac{12 x}{x^{2} + 8} + \frac{6 {\color{red}\left(\frac{d}{dx} \left(x\right)\right)}}{x + 1} = - \frac{12 x}{x^{2} + 8} + \frac{6 {\color{red}\left(1\right)}}{x + 1}$$

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

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

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