Question

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

Answer 1

`dy/dx = (1-ln(x))/(x^2ln(10))`

Step by step explanation is given below

Explanation:

Interesting question! To find derivative of `y=log_10(x)/x`

Most people would be confused because of the `log_10(x)`

We know `d/dx(ln(x)) = 1/x`

So let us make `log_10(x)` 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(a) = ln(a)/ln(b)`

Now using this with our `log_10(x)`
We can write it as `ln(x)/ln(10)` This we can work with.

Our `y=log_10(x)/x`

`y=ln(x)/(ln(10)x)`

We can use the quotient rule to simplify this.

`color(red)("Quotient rule :"` `(u/v)'=(vu'-v'u)/v^2`

`"Let "u=ln(x)/ln(10) quad " and "quad quad quad v=x`

Differentiating with respect to `x`

`u' = 1/(xln(10)) " and " v' = 1`

Now we find the derivative

`dy/dx = (x(1/(xln(10))-ln(x)/ln(10)*1)) /x^2`

`dy/dx = (1/ln(10)-ln(x)/ln(10)) /x^2`

`dy/dx = (1-ln(x))/(x^2ln(10))`