# How do you find the derivative of y= log _ 10 x/x?

Jan 11, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - \ln \left(x\right)}{{x}^{2} \ln \left(10\right)}$

Step by step explanation is given below

#### Explanation:

Interesting question! To find derivative of $y = {\log}_{10} \frac{x}{x}$

Most people would be confused because of the ${\log}_{10} \left(x\right)$

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

So let us make ${\log}_{10} \left(x\right)$ into something which we know.

Here the change of base of rule would come in handy.

color(red)("Change of base rule" quad ${\log}_{b} \left(a\right) = \ln \frac{a}{\ln} \left(b\right)$

Now using this with our ${\log}_{10} \left(x\right)$
We can write it as $\ln \frac{x}{\ln} \left(10\right)$ This we can work with.

Our $y = {\log}_{10} \frac{x}{x}$

$y = \ln \frac{x}{\ln \left(10\right) x}$

We can use the quotient rule to simplify this.

color(red)("Quotient rule :" $\left(\frac{u}{v}\right) ' = \frac{v u ' - v ' u}{v} ^ 2$

$\text{Let "u=ln(x)/ln(10) quad " and } \quad \quad \quad v = x$

Differentiating with respect to $x$

$u ' = \frac{1}{x \ln \left(10\right)} \text{ and } v ' = 1$

Now we find the derivative

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x \left(\frac{1}{x \ln \left(10\right)} - \ln \frac{x}{\ln} \left(10\right) \cdot 1\right)}{x} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{1}{\ln} \left(10\right) - \ln \frac{x}{\ln} \left(10\right)}{x} ^ 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1 - \ln \left(x\right)}{{x}^{2} \ln \left(10\right)}$