basically, doing logarithmic differntiation is just to use log rules to simplify the calculation of differentiation. You may ask why we need to simplify the calculation, but you will understand why after I show you an example below.

If I ask you to find the dirivative of y= `sqrt(x) (e^ (x^2)) * ((x^2) +1)^10`

What are you going to do?

If we don't use logarithmic differtiation to simplify the equation first, we will have to use product rule for multiple times and other rules as well for multiple times. I am sure you don't have the time to do this thing in the exam you are doing it that way. Besides, more calculation means the possibility of making mistakes is higher. So, we need logarithmic differentiation. But before that, let's review some log rules:

ln (xy)=lnx + lny

ln(x/y)=lnx - lny

ln(x^r)= r ln x

So now we are ready to go:

lny= ln (`sqrt(x) (e^ (x^2)) * ((x^2) +1)^10`)

lny=ln sqrt(x) + ln (e^ (x^2)) + ln (((x^2) +1)^10)

lny=ln (x^(1/2)) + ln (e^ (x^2)) + ln (((x^2) +1)^10)

lny= (1/2) ln x + x^2 * ln e + 10 ln (x^2 +1)

lny= (1/2) ln x + x^2 * 1 + 10 ln (x^2 +1)

lny= (1/2) ln x + x^2 + 10 ln (x^2 +1)

Now we are going to do the differtiation:

y' / y = (1/2) (1/x) + 2x + 10 ( 1/ (x^2 + 1 )) (2x)

y' = (1/2) (1/x) + 2x + 10 ( 1/ (x^2 + 1 )) (2x) * y

And since y=`sqrt(x) (e^ (x^2)) * ((x^2) +1)^10``

We get:

y' = (1/2) (1/x) + 2x + 10 ( 1/ (x^2 + 1 )) (2x) * (`sqrt(x) (e^ (x^2)) * ((x^2) +1)^10)

You now see you efficient logarithmic differtiation is? That saves you a lot of time.

**Subject :**Math**Topic :**Calculus-
**Posted By :**Jason