# How do you find the derivative of y= log_4(x+e^x+e)?

Dec 28, 2016

$= \frac{1}{\ln} 4 \left(\frac{1 + {e}^{x}}{x + {e}^{x} + 1}\right)$

#### Explanation:

Use ${\log}_{b} a = {\log}_{c} \frac{a}{\log} _ c b \mathmr{and} \left(\ln u\right) ' = \left(\frac{1}{u}\right) u '$

So,

$y ' = \left(\ln \frac{x + {e}^{x} + e}{\ln} 4\right) '$

$= \frac{\frac{1}{x + {e}^{x} + e} \left(x + {e}^{x} + e\right) '}{\ln} 4$

$= \frac{1}{\ln} 4 \left(\frac{1 + {e}^{x}}{x + {e}^{x} + 1}\right)$