# How do you differentiate f(x) = log_3(log_7 (log_5(x + 2)))?

Jun 8, 2016

$f ' \left(x\right) = \frac{1}{\ln \left(3\right) \left(x + 2\right) \ln \left({\log}_{5} \left(x + 2\right)\right) \ln \left(x + 2\right)}$

#### Explanation:

The typical method for differentiating a logarithm that doesn't have base $e$ requires using the change of base formula to put the expression into terms of the natural logarithm.

For example, the derivative of log_a(g(x))) can be found through saying that

$\frac{d}{\mathrm{dx}} {\log}_{a} \left(g \left(x\right)\right) = \frac{d}{\mathrm{dx}} \left(\ln \frac{g \left(x\right)}{\ln} \left(a\right)\right) = \frac{1}{\ln} \left(a\right) \left(\frac{d}{\mathrm{dx}} \ln \left(g \left(x\right)\right)\right)$

$= \frac{1}{\ln} \left(a\right) \left(\frac{1}{g} \left(x\right)\right) \cdot g ' \left(x\right) = \frac{1}{g \left(x\right) \ln \left(a\right)} \cdot g ' \left(x\right)$

So, this rule will give us:

$f ' \left(x\right) = \frac{1}{{\log}_{7} \left({\log}_{5} \left(x + 2\right)\right) \ln \left(3\right)} \cdot \frac{d}{\mathrm{dx}} {\log}_{7} \left({\log}_{5} \left(x + 2\right)\right)$

Using this rule again on the smaller logarithm function (like peeling off its shells):

$f ' \left(x\right) = \frac{1}{{\log}_{7} \left({\log}_{5} \left(x + 2\right)\right) \ln \left(3\right)} \left(\frac{1}{{\log}_{5} \left(x + 2\right) \ln \left(7\right)}\right) \cdot \frac{d}{\mathrm{dx}} {\log}_{5} \left(x + 2\right)$

Noting that $\frac{d}{\mathrm{dx}} {\log}_{5} \left(x + 2\right) = \frac{1}{\left(x + 2\right) \ln \left(5\right)} \cdot \frac{d}{\mathrm{dx}} \left(x + 2\right)$, which equals $\frac{1}{\left(x + 2\right) \ln \left(5\right)}$, the entire function's derivative equals

$f ' \left(x\right) = \frac{1}{\ln \left(3\right) \ln \left(5\right) \ln \left(7\right) \left(x + 2\right) {\log}_{7} \left({\log}_{5} \left(x + 2\right)\right) {\log}_{5} \left(x + 2\right)}$

This can be simplified by using the change of base formula on ${\log}_{7} \left({\log}_{5} \left(x + 2\right)\right)$ and log_5(x+2)):

$f ' \left(x\right) = \frac{1}{\ln \left(3\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{\ln \left(5\right)}}} \textcolor{red}{\cancel{\textcolor{b l a c k}{\ln \left(7\right)}}} \left(x + 2\right) \left(\ln \frac{{\log}_{5} \left(x + 2\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\ln \left(7\right)}}}}\right) \left(\ln \frac{x + 2}{\textcolor{red}{\cancel{\textcolor{b l a c k}{\ln \left(5\right)}}}}\right)}$

$f ' \left(x\right) = \frac{1}{\ln \left(3\right) \left(x + 2\right) \ln \left({\log}_{5} \left(x + 2\right)\right) \ln \left(x + 2\right)}$