# How do you find the derivative y=ln(9x)/(1+x)?

Mar 1, 2015

The answer is: $y ' = \frac{1 + x - x \ln 9 x}{x {\left(1 + x\right)}^{2}}$.

Using the quotient rule and the chain rule:

$y ' = \frac{\frac{1}{9 x} \cdot 9 \cdot \left(1 + x\right) - \ln 9 x \cdot 1}{1 + x} ^ 2 = \frac{\frac{1 + x}{x} - \ln 9 x}{1 + x} ^ 2 =$

$= \frac{1 + x - x \ln 9 x}{x {\left(1 + x\right)}^{2}}$.