Afgeleide van $$$\left(x^{3} + 2\right)^{2} \left(x^{4} + 4\right)^{4}$$$

Zij $$$H{\left(x \right)} = \left(x^{3} + 2\right)^{2} \left(x^{4} + 4\right)^{4}$$$.

Neem de logaritme van beide zijden: $$$\ln\left(H{\left(x \right)}\right) = \ln\left(\left(x^{3} + 2\right)^{2} \left(x^{4} + 4\right)^{4}\right)$$$.

Herschrijf het rechterlid met behulp van de eigenschappen van logaritmen: $$$\ln\left(H{\left(x \right)}\right) = 2 \ln\left(x^{3} + 2\right) + 4 \ln\left(x^{4} + 4\right)$$$.

Differentieer afzonderlijk beide zijden van de vergelijking: $$$\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right) = \frac{d}{dx} \left(2 \ln\left(x^{3} + 2\right) + 4 \ln\left(x^{4} + 4\right)\right)$$$.

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:

$${\color{red}\left(\frac{d}{dx} \left(2 \ln\left(x^{3} + 2\right) + 4 \ln\left(x^{4} + 4\right)\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(2 \ln\left(x^{3} + 2\right)\right) + \frac{d}{dx} \left(4 \ln\left(x^{4} + 4\right)\right)\right)}$$

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 $$$c = 2$$$ en $$$f{\left(x \right)} = \ln\left(x^{3} + 2\right)$$$:

$${\color{red}\left(\frac{d}{dx} \left(2 \ln\left(x^{3} + 2\right)\right)\right)} + \frac{d}{dx} \left(4 \ln\left(x^{4} + 4\right)\right) = {\color{red}\left(2 \frac{d}{dx} \left(\ln\left(x^{3} + 2\right)\right)\right)} + \frac{d}{dx} \left(4 \ln\left(x^{4} + 4\right)\right)$$

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 $$$c = 4$$$ en $$$f{\left(x \right)} = \ln\left(x^{4} + 4\right)$$$:

$${\color{red}\left(\frac{d}{dx} \left(4 \ln\left(x^{4} + 4\right)\right)\right)} + 2 \frac{d}{dx} \left(\ln\left(x^{3} + 2\right)\right) = {\color{red}\left(4 \frac{d}{dx} \left(\ln\left(x^{4} + 4\right)\right)\right)} + 2 \frac{d}{dx} \left(\ln\left(x^{3} + 2\right)\right)$$

De functie $$$\ln\left(x^{3} + 2\right)$$$ is de samenstelling $$$f{\left(g{\left(x \right)} \right)}$$$ van twee functies $$$f{\left(u \right)} = \ln\left(u\right)$$$ en $$$g{\left(x \right)} = x^{3} + 2$$$.

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:

$$2 {\color{red}\left(\frac{d}{dx} \left(\ln\left(x^{3} + 2\right)\right)\right)} + 4 \frac{d}{dx} \left(\ln\left(x^{4} + 4\right)\right) = 2 {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(x^{3} + 2\right)\right)} + 4 \frac{d}{dx} \left(\ln\left(x^{4} + 4\right)\right)$$

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

$$2 {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(x^{3} + 2\right) + 4 \frac{d}{dx} \left(\ln\left(x^{4} + 4\right)\right) = 2 {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(x^{3} + 2\right) + 4 \frac{d}{dx} \left(\ln\left(x^{4} + 4\right)\right)$$

Keer terug naar de oorspronkelijke variabele:

$$4 \frac{d}{dx} \left(\ln\left(x^{4} + 4\right)\right) + \frac{2 \frac{d}{dx} \left(x^{3} + 2\right)}{{\color{red}\left(u\right)}} = 4 \frac{d}{dx} \left(\ln\left(x^{4} + 4\right)\right) + \frac{2 \frac{d}{dx} \left(x^{3} + 2\right)}{{\color{red}\left(x^{3} + 2\right)}}$$

De functie $$$\ln\left(x^{4} + 4\right)$$$ is de samenstelling $$$f{\left(g{\left(x \right)} \right)}$$$ van twee functies $$$f{\left(u \right)} = \ln\left(u\right)$$$ en $$$g{\left(x \right)} = x^{4} + 4$$$.

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:

$$4 {\color{red}\left(\frac{d}{dx} \left(\ln\left(x^{4} + 4\right)\right)\right)} + \frac{2 \frac{d}{dx} \left(x^{3} + 2\right)}{x^{3} + 2} = 4 {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(x^{4} + 4\right)\right)} + \frac{2 \frac{d}{dx} \left(x^{3} + 2\right)}{x^{3} + 2}$$

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

$$4 {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(x^{4} + 4\right) + \frac{2 \frac{d}{dx} \left(x^{3} + 2\right)}{x^{3} + 2} = 4 {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(x^{4} + 4\right) + \frac{2 \frac{d}{dx} \left(x^{3} + 2\right)}{x^{3} + 2}$$

Keer terug naar de oorspronkelijke variabele:

$$\frac{4 \frac{d}{dx} \left(x^{4} + 4\right)}{{\color{red}\left(u\right)}} + \frac{2 \frac{d}{dx} \left(x^{3} + 2\right)}{x^{3} + 2} = \frac{4 \frac{d}{dx} \left(x^{4} + 4\right)}{{\color{red}\left(x^{4} + 4\right)}} + \frac{2 \frac{d}{dx} \left(x^{3} + 2\right)}{x^{3} + 2}$$

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

$$\frac{4 {\color{red}\left(\frac{d}{dx} \left(x^{4} + 4\right)\right)}}{x^{4} + 4} + \frac{2 \frac{d}{dx} \left(x^{3} + 2\right)}{x^{3} + 2} = \frac{4 {\color{red}\left(\frac{d}{dx} \left(x^{4}\right) + \frac{d}{dx} \left(4\right)\right)}}{x^{4} + 4} + \frac{2 \frac{d}{dx} \left(x^{3} + 2\right)}{x^{3} + 2}$$

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

$$\frac{4 \left({\color{red}\left(\frac{d}{dx} \left(4\right)\right)} + \frac{d}{dx} \left(x^{4}\right)\right)}{x^{4} + 4} + \frac{2 \frac{d}{dx} \left(x^{3} + 2\right)}{x^{3} + 2} = \frac{4 \left({\color{red}\left(0\right)} + \frac{d}{dx} \left(x^{4}\right)\right)}{x^{4} + 4} + \frac{2 \frac{d}{dx} \left(x^{3} + 2\right)}{x^{3} + 2}$$

Pas de machtsregel $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ toe met $$$n = 4$$$:

$$\frac{4 {\color{red}\left(\frac{d}{dx} \left(x^{4}\right)\right)}}{x^{4} + 4} + \frac{2 \frac{d}{dx} \left(x^{3} + 2\right)}{x^{3} + 2} = \frac{4 {\color{red}\left(4 x^{3}\right)}}{x^{4} + 4} + \frac{2 \frac{d}{dx} \left(x^{3} + 2\right)}{x^{3} + 2}$$

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

$$\frac{16 x^{3}}{x^{4} + 4} + \frac{2 {\color{red}\left(\frac{d}{dx} \left(x^{3} + 2\right)\right)}}{x^{3} + 2} = \frac{16 x^{3}}{x^{4} + 4} + \frac{2 {\color{red}\left(\frac{d}{dx} \left(x^{3}\right) + \frac{d}{dx} \left(2\right)\right)}}{x^{3} + 2}$$

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

$$\frac{16 x^{3}}{x^{4} + 4} + \frac{2 \left({\color{red}\left(\frac{d}{dx} \left(2\right)\right)} + \frac{d}{dx} \left(x^{3}\right)\right)}{x^{3} + 2} = \frac{16 x^{3}}{x^{4} + 4} + \frac{2 \left({\color{red}\left(0\right)} + \frac{d}{dx} \left(x^{3}\right)\right)}{x^{3} + 2}$$

Pas de machtsregel $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ toe met $$$n = 3$$$:

$$\frac{16 x^{3}}{x^{4} + 4} + \frac{2 {\color{red}\left(\frac{d}{dx} \left(x^{3}\right)\right)}}{x^{3} + 2} = \frac{16 x^{3}}{x^{4} + 4} + \frac{2 {\color{red}\left(3 x^{2}\right)}}{x^{3} + 2}$$

Dus, $$$\frac{d}{dx} \left(2 \ln\left(x^{3} + 2\right) + 4 \ln\left(x^{4} + 4\right)\right) = \frac{16 x^{3}}{x^{4} + 4} + \frac{6 x^{2}}{x^{3} + 2}$$$.

Dus, $$$\frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{H{\left(x \right)}} = \frac{16 x^{3}}{x^{4} + 4} + \frac{6 x^{2}}{x^{3} + 2}$$$.

Daarom geldt $$$\frac{d}{dx} \left(H{\left(x \right)}\right) = \left(\frac{16 x^{3}}{x^{4} + 4} + \frac{6 x^{2}}{x^{3} + 2}\right) H{\left(x \right)} = 2 x^{2} \left(x^{3} + 2\right) \left(x^{4} + 4\right)^{3} \left(3 x^{4} + 8 x \left(x^{3} + 2\right) + 12\right).$$$