Derivative of $$$\left(x^{2} + 2\right)^{2} \left(x^{4} + 4\right)^{4}$$$

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

Take the logarithm of both sides: $$$\ln\left(H{\left(x \right)}\right) = \ln\left(\left(x^{2} + 2\right)^{2} \left(x^{4} + 4\right)^{4}\right)$$$.

Rewrite the RHS using the properties of logarithms: $$$\ln\left(H{\left(x \right)}\right) = 2 \ln\left(x^{2} + 2\right) + 4 \ln\left(x^{4} + 4\right)$$$.

Differentiate separately both sides of the equation: $$$\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right) = \frac{d}{dx} \left(2 \ln\left(x^{2} + 2\right) + 4 \ln\left(x^{4} + 4\right)\right)$$$.

Differentiate the LHS of the equation.

The function $$$\ln\left(H{\left(x \right)}\right)$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = \ln\left(u\right)$$$ and $$$g{\left(x \right)} = H{\left(x \right)}$$$.

Apply the chain rule $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

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

The derivative of the natural logarithm is $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:

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

Return to the old variable:

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

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

Differentiate the RHS of the equation.

The derivative of a sum/difference is the sum/difference of derivatives:

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

Apply the constant multiple rule $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ with $$$c = 4$$$ and $$$f{\left(x \right)} = \ln\left(x^{4} + 4\right)$$$:

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

The function $$$\ln\left(x^{4} + 4\right)$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = \ln\left(u\right)$$$ and $$$g{\left(x \right)} = x^{4} + 4$$$.

Apply the chain rule $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

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

The derivative of the natural logarithm is $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:

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

Return to the old variable:

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

The derivative of a sum/difference is the sum/difference of derivatives:

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

Apply the constant multiple rule $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ with $$$c = 2$$$ and $$$f{\left(x \right)} = \ln\left(x^{2} + 2\right)$$$:

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

The function $$$\ln\left(x^{2} + 2\right)$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = \ln\left(u\right)$$$ and $$$g{\left(x \right)} = x^{2} + 2$$$.

Apply the chain rule $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:

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

The derivative of the natural logarithm is $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:

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

Return to the old variable:

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

The derivative of a sum/difference is the sum/difference of derivatives:

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

The derivative of a constant is $$$0$$$:

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

Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = 2$$$:

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

The derivative of a constant is $$$0$$$:

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

Apply the power rule $$$\frac{d}{dx} \left(x^{n}\right) = n x^{n - 1}$$$ with $$$n = 4$$$:

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

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

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

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