00:01
Hello students we are going to write here y value is l x l and x raise to power 1 upon l and x we are going to find the derivative of y so we are going to write here y is equal to l n x raise to power 1 upon ln x so basically if i write here for this so take log of both sides log of both sides we are going to write here law of both sides we are going to write here for this of both sides taking base as e taking base as e we are going to write here for this this must be ln of y is equivalent to ln of x raised to power we are going to write here for this one of one divided by l n of x so basically we are going to write here for this so this must be written something like this we can write here ln of a raise to power x is x ln of a so this must be same we need to apply the formula 1 upon ln of x and this is ln of x so now we are going to calculate it let differentiate both sides.
01:27
So differentiate, we are going to write here for this.
01:32
Differentiate both sides.
01:35
So differentiation of lhs must be equivalent to chain rule, which is 1 upon by d .y of dx.
01:43
We are going to write here, differentiation of ln of x is 1 upon x.
01:47
So this must be differentiated by u by v method, where differentiation of u.
01:55
Divided by v is equivalent to v is square and this must be view then differentiation of u minus u and differentiation of v so we are going to write here for this ln of x raise to power two we are going to write here for this ln of s is same then one upon of l n x this is differentiated and then one upon x and minus 1 upon of x so ln of ln x so basically we are going to extend this to make it more clear so let's erase this to make it to make it more clear now d y y t x is equivalent to y y value is ln of x raise to power 1 upon ln of x then this this must be written as ln of x.
02:54
So we are going to write here, this is ln of x.
02:58
So this must be erased to make it more clear because this is in whole...