Find $$ y^{\prime} $$ $$ y=\left(\frac{1}{x}+7\right)\left(x-\frac{1}{x}+7\right) $$

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Find the first order derivative with respect to $$ x $$ for $$ y=(\frac{1}{x}+7)(x-\frac{1}{x}+7) $$.
Evaluate the derivative by following steps:
- step0: Evaluate the derivative:
  $$y^{\prime}=\frac{d}{dx}\left(\frac{\left(1+7xight)\left(x^{2}-1+7xight)}{x^{2}}ight)$$
- step1: Use differentiation rules:
  $$y^{\prime}=\frac{\frac{d}{dx}\left(\left(1+7xight)\left(x^{2}-1+7xight)ight)	imes x^{2}-\left(1+7xight)\left(x^{2}-1+7xight)	imes \frac{d}{dx}\left(x^{2}ight)}{\left(x^{2}ight)^{2}}$$
- step2: Calculate:
  $$y^{\prime}=\frac{\left(21x^{2}+100xight)x^{2}-\left(1+7xight)\left(x^{2}-1+7xight)	imes 2x}{\left(x^{2}ight)^{2}}$$
- step3: Multiply the terms:
  $$y^{\prime}=\frac{x^{2}\left(21x^{2}+100xight)-2\left(1+7xight)\left(x^{2}-1+7xight)x}{x^{4}}$$
- step4: Subtract the terms:
  $$y^{\prime}=\frac{7x^{4}+2x}{x^{4}}$$
- step5: Divide the terms:
  $$y^{\prime}=\frac{7x^{3}+2}{x^{3}}$$
The derivative of $$ y $$ with respect to $$ x $$ is $$ y' = \frac{7x^3 + 2}{x^3} $$.
$$ y' = \frac{7x^3 + 2}{x^3} $$

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