# What is the derivative of y=log_10(x)?

Jul 25, 2014

$y ' = {\log}_{10} \left(e\right) \cdot \frac{1}{x}$

Solution

Suppose we have ${\log}_{a} \left(b\right)$, we want to change it on exponential (e) base, then it can be written as:

${\log}_{a} \left(b\right) = {\log}_{a} \left(e\right) \cdot {\log}_{e} \left(b\right)$

Similarly, function ${\log}_{10} \left(x\right)$ can be written as:

$y = {\log}_{10} \left(e\right) \cdot {\log}_{e} \left(x\right)$

Let's say we have, $y = c \cdot f \left(x\right)$, where c is a constant
then, $y ' = c \cdot f ' \left(x\right)$

Now, this is quite straightforward to differentiate, as ${\log}_{10} \left(e\right)$ is constant, so only remaining function is ${\log}_{e} \left(x\right)$

Hence:

$y ' = {\log}_{10} \left(e\right) \cdot \frac{1}{x}$

Alternate solution:

Another common approach is to use the change of base formula, which says that:

${\log}_{a} \left(b\right) = \ln \frac{b}{\ln} \left(a\right)$

From change of base we have ${\log}_{10} \left(x\right) = {\log}_{10} \left(x\right) = \ln \frac{x}{\ln} \left(10\right)$.

This we can differentiate as long as we remember that

$\frac{1}{\ln} \left(10\right)$ is just a constant multipler.

Doing the problem this way gives a result of $y ' = \frac{1}{\ln} \left(10\right) \cdot \frac{1}{x}$.