# How do I find the derivative of 15/2+ln(5x) ?

Jan 9, 2016

$\frac{1}{x}$

#### Explanation:

First, notice that differentiating the $\frac{15}{2}$ will just give $0$, so this question is identical to just finding the derivative of $\ln \left(5 x\right)$.

In order to differentiate functions with the natural logarithm, it's necessary to know that $\frac{d}{\mathrm{dx}} \left(\ln \left(x\right)\right) = \frac{1}{x}$.

Then, through the chain rule, this can be generalized to say that $\frac{d}{\mathrm{dx}} \left(\ln \left(u\right)\right) = \frac{1}{u} \cdot u '$.

Thus,

$\frac{d}{\mathrm{dx}} \left(\ln \left(5 x\right)\right) = \frac{1}{5 x} \cdot \frac{d}{\mathrm{dx}} \left(5 x\right)$

Since $\frac{d}{\mathrm{dx}} \left(5 x\right) = 5$, the derivative of the original function is

$\frac{1}{5 x} \cdot 5 = \frac{1}{x}$

Notice that this derivative is the exact same as the derivative of just $\ln \left(x\right)$. This can be explained using logarithm rules:

$\ln \left(5 x\right) = \ln \left(5\right) + \ln \left(x\right)$

So, $\frac{d}{\mathrm{dx}} \left(\ln \left(5\right) + \ln \left(x\right)\right) = 0 + \frac{1}{x} = \frac{1}{x}$