Question

differentiate y=x^x, x>0

Answer 1

take log on both sides

Answer 2

log y=x log(x)
apply product rule on rhs

Answer 3

any doubt??

Answer 4

taking log or ln?

Answer 5

natural log ln.......

Answer 6

then i get this answer so far.. 1/y=ln x.. then what?

Answer 7

you forgot to use the chain rule on the left side...

Answer 8

what chain rule, isn't it only on the right hand side?

Answer 9

the first step is : ln y=x ln x right?

Answer 10

use the product rule on the right...

Answer 11

yes... take the derivative with respect to x on both sides.

Answer 12

then i get this.. 1/y=ln x.. then what?

Answer 13

you still didn't differentiate that left side properly.

Answer 14

how?

Answer 15

\(\large [lnx]'=\frac{1}{x} \) , correct?

so what is

\(\large [lny]'= \) ??? use the chain rule...

Answer 16

"y" is a function of x....

Answer 17

omg, my calculus is terrible.. haha.. does this related to implicit diff? dy/dx(1/x)?

Answer 18

well we are differentiating implicitly the relation: lny=xlnx

Answer 19

here is the chain rule formula: \(\large [lnf(x)]'=\frac{1}{f(x)}\cdot f'(x)=\frac{f'(x)}{f(x)} \)

Answer 20

i found the solution already... :P http://www.analyzemath.com/calculus/Differentiation/first_derivative.html tq:)

Answer 21

glad u found it... as you can see from the solution you found, [lny]' = y'/y... what i was trying to tell u....