How do you find the derivative of y=ln(5x)?

Jul 29, 2014

$y ' = \frac{1}{x}$

Explanation

Suppose, $y = \ln \left(b \left(x\right)\right)$

then using Chain Rule,

$y ' = \frac{1}{b \left(x\right)} \cdot \left(b \left(x\right)\right) '$

Similarly following for the above function yields,

$y ' = \frac{1}{5 x} \cdot \left(5 x\right) '$

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

$y ' = \frac{1}{x}$

Mar 14, 2015

A student comfortable with the natural logarithm function and its properties might think of this:

One could reason as follows:
$y = \ln \left(5 x\right) = \ln \left(5\right) + \ln \left(x\right)$.

But $\ln \left(5\right)$ is a constant, so its derivative is $0$.

Therefore, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left(\ln 5 + \ln x\right) = \frac{d}{\mathrm{dx}} \left(\ln x\right) = \frac{1}{x}$.

Mar 24, 2015

Using the Chain Rule you'll get:

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

(Derive $\ln$ as it is and then multiply by the derivative of the argument, $5 x$).