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La derivada será
![y = (x + ln(x) )^{5} \\ \\ y' = 5 {(x + ln(x) )}^{4} \times (x + ln(x) )' \\ \\ y' = 5 {(x + ln(x) )}^{4} \times (x' + ln(x) ') \\ \\ y' = 5 {(x + ln(x) )}^{4} \times(1 + \frac{1}{x} ) \\ \\ y' = \frac{5 {(x + ln(x) }^{4} )(x + 1)}{x} y = (x + ln(x) )^{5} \\ \\ y' = 5 {(x + ln(x) )}^{4} \times (x + ln(x) )' \\ \\ y' = 5 {(x + ln(x) )}^{4} \times (x' + ln(x) ') \\ \\ y' = 5 {(x + ln(x) )}^{4} \times(1 + \frac{1}{x} ) \\ \\ y' = \frac{5 {(x + ln(x) }^{4} )(x + 1)}{x}](https://tex.z-dn.net/?f=y+%3D+%28x+%2B++ln%28x%29+%29%5E%7B5%7D++%5C%5C++%5C%5C+y%27++%3D+5+%7B%28x+%2B++ln%28x%29+%29%7D%5E%7B4%7D++%5Ctimes+%28x+%2B++ln%28x%29+%29%27+%5C%5C++%5C%5C++y%27++%3D+5+%7B%28x+%2B++ln%28x%29+%29%7D%5E%7B4%7D++%5Ctimes+%28x%27+%2B++ln%28x%29+%27%29+%5C%5C++%5C%5C++y%27++%3D+5+%7B%28x+%2B++ln%28x%29+%29%7D%5E%7B4%7D++%5Ctimes%281+%2B++%5Cfrac%7B1%7D%7Bx%7D+%29+%5C%5C++%5C%5C+y%27+%3D++%5Cfrac%7B5+%7B%28x+%2B++ln%28x%29+%7D%5E%7B4%7D+%29%28x+%2B+1%29%7D%7Bx%7D+)
La derivada será
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