Question

find the derivative of y=x^x in two different ways
i got 1 way down by making x^x=xx = ex*ln(x) and just taking it from there...

Answer 1

...honestly idk what you did haha

Answer 2

but here's one way (idk if this is what you did)
let y = x^x

ln both sides
\[\ln y = x \ln x\]implicitly differentiate
\[\frac 1y y' = 1 + \ln x\]
corss multiply\[y' = y(1 + \ln x)\]
\[y' = x^x (1 + \ln x)\]

Answer 3

another way:

let x^x= y

take natural log on both sides

x (ln x) = ln y

Now, differentiate both sides w.r.t. x [use product rule for LHS and for RHS use chain rule]
I hope that makes sense..

Answer 4

ya i thought of differentiation too :o

Answer 5

@lgbasallote hit the nail on the head... That is it..

Answer 6

is that what you did @alexwee123 ?

Answer 7

uh ya kinda but i got confused for some reason and never rlly did it :/but now i get it :D

Answer 8

No, @alexwee123 did is:

x^x = e^ ln(x^x) = e^(x lnx)

Now, he differentiated using the long chain rule

Answer 9

hmm i didnt know about that :o

Answer 10

that looks fun

Answer 11

Suggestion Please!!!

I am studying the Multivariable Calculus from MIT Open Courseware lectures, yes the course 18.02.. and I feel like not understanding the topics fully.

can anyone suggest me some better course on Multivariable Calculus. I googled an d saw that UC Berkeley has a course on that. Is that better than that of MIT?

Answer 12

@lgbasallote I would appreciate a response from you.. for some reason system isn't letting me send you a message...

Answer 13

really? it should....anyway...im not really familiar with websites...im very naive in the net hah...bottomline idk any website

Answer 14

or any US school

Answer 15

Okay.. no probs