Derivada de $$$\left(x^{3} + 2\right)^{2} \left(x^{4} + 4\right)^{4}$$$

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

Pegue o logaritmo de ambos os lados: $$$\ln\left(H{\left(x \right)}\right) = \ln\left(\left(x^{3} + 2\right)^{2} \left(x^{4} + 4\right)^{4}\right)$$$.

Reescreva o RHS usando as propriedades dos logaritmos: $$$\ln\left(H{\left(x \right)}\right) = 2 \ln\left(x^{3} + 2\right) + 4 \ln\left(x^{4} + 4\right)$$$.

Diferencie separadamente os dois lados da equação: $$$\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right) = \frac{d}{dx} \left(2 \ln\left(x^{3} + 2\right) + 4 \ln\left(x^{4} + 4\right)\right)$$$.

Diferencie o LHS da equação.

A função $$$\ln\left(H{\left(x \right)}\right)$$$ é a composição $$$f{\left(g{\left(x \right)} \right)}$$$ de duas funções $$$f{\left(u \right)} = \ln\left(u\right)$$$ e $$$g{\left(x \right)} = H{\left(x \right)}$$$.

Aplique a regra da cadeia $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

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

A derivada do logaritmo natural é $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:

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

Volte para a variável antiga:

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

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

Diferencie o RHS da equação.

A derivada de uma soma/diferença é a soma/diferença das derivadas:

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

Aplique a regra múltipla constante $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ com $$$c = 4$$$ e $$$f{\left(x \right)} = \ln\left(x^{4} + 4\right)$$$:

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

Aplique a regra múltipla constante $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ com $$$c = 2$$$ e $$$f{\left(x \right)} = \ln\left(x^{3} + 2\right)$$$:

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

A função $$$\ln\left(x^{4} + 4\right)$$$ é a composição $$$f{\left(g{\left(x \right)} \right)}$$$ de duas funções $$$f{\left(u \right)} = \ln\left(u\right)$$$ e $$$g{\left(x \right)} = x^{4} + 4$$$.

Aplique a regra da cadeia $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

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

A derivada do logaritmo natural é $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:

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

Volte para a variável antiga:

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

A derivada de uma soma/diferença é a soma/diferença das derivadas:

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

Aplique a regra de poder $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ com $$$n = 4$$$:

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

A derivada de uma constante é $$$0$$$:

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

A função $$$\ln\left(x^{3} + 2\right)$$$ é a composição $$$f{\left(g{\left(x \right)} \right)}$$$ de duas funções $$$f{\left(u \right)} = \ln\left(u\right)$$$ e $$$g{\left(x \right)} = x^{3} + 2$$$.

Aplique a regra da cadeia $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

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

A derivada do logaritmo natural é $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:

$$\frac{16 x^{3}}{x^{4} + 4} + 2 {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(x^{3} + 2\right) = \frac{16 x^{3}}{x^{4} + 4} + 2 {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(x^{3} + 2\right)$$

Volte para a variável antiga:

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

A derivada de uma soma/diferença é a soma/diferença das derivadas:

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

A derivada de uma constante é $$$0$$$:

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

Aplique a regra de poder $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ com $$$n = 3$$$:

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

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

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

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