Expert: Paul Klarreich - 11/20/2009

Question
Hi Paul,
1. Pls help me with the Taylor series expansion around x=2 of y=LOG5 [3^(x+1)]
2. How to solve the derivative of y function y'(x)
where y=x^x

Thank you for your help.  

Answer
Questioner: Geli
Country: United States
Category: Advanced Math
Private: No
Subject: Taylor series and derivative
Question: Hi Paul,
1. Pls help me with the Taylor series expansion around x=2 of y=LOG5 [3^(x+1)]
2. How to solve the derivative of y function y'(x)
where y=x^x

Thank you for your help.
.....................................

If y=LOG5 [3^(x+1)]

recall that  log[b] (x) = log[b](a) log[a](x), so:


LOG5 [3^(x+1)] =

log5(3) log3(3^(x+1)) =

log5(3) TIMES (x+1),

and log5(3) is simply a constant, which I will call L.

So y = L (x+1).

y = L (x - 2 + 3)

y = L[3 +  (x - 2) ]

and that looks like a Taylor series to me, even if it is not very long.


2. How to solve the derivative of y function y'(x)
where y=x^x

Your vocabulary does not make sense.  Do you mean:

If  y = x^x,  compute  y'?

If so, just write:

y = (e ^ ln x)^x = e^(x ln x)

Now use the chain rule, then the product rule:

y' = e^(x ln x)(ln x + 1)

y' = x^x (ln x + 1)

Note;  Being careless with the vocabulary is a good way to fail math.  Start learning the vocabulary and using it correctly -- you will do a lot better.