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

Jan 28, 2015

The quotient rule states that, given an expression $f \left(x\right) = g \frac{x}{h \left(x\right)}$, the derivative $f ' \left(x\right) = \frac{g ' \left(x\right) h \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2$

In the case above, our functions are:

$g \left(x\right) = \ln \left(9 x\right)$ and $h \left(x\right) = 1 + x$.

Giving us derivatives of:

$g ' \left(x\right) = \left(\frac{d}{\mathrm{dx}}\right) \ln \left(9 x\right) = \frac{9}{9 x} = \frac{1}{x}$ and $h ' \left(x\right) = \left(\frac{d}{\mathrm{dx}}\right) \left(1 + x\right) = 1$

Thus, by the Quotient Rule:

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

Note that $g \left(x\right)$ is undefined in the real number plane for any $x \le 0$; therefore, the derivative will be undefined as well (recall that $\ln \left(x\right)$ does not have a real value for $x \le 0$). Thus, since the only point at which our denominator is 0 would be $x = - 1$, we need not worry about the denominator causing discontinuities within the domain of our function $f \left(x\right)$.