Expert: Paul Klarreich - 1/25/2010

Question
calculus 1
y=x^(sin(x))
ln(y)=sin(x)ln(x)
y'/y=sin(x)/x+ ln(x)cos(x)
y'=x^sin(x)(sin(x)/x +ln(x)cos(x))
now I am having problems
x^sin(x) never =0 and its undefined at x =0 but since its the same original function its also undefined so I cant find a critical point
sin(x)/x + ln(x)cos(x)=0
sin(x)/x = -ln(x)cos(x)
tan(x) = -ln(x)x
I don't know enough trigonometry to solve this  

Answer
Questioner: hamad
Country: Kuwait
Category: Calculus
Private: No
Subject: critical point
Question: calculus 1

y=x^(sin(x))

ln(y)=sin(x)ln(x)

y'/y=sin(x)/x+ ln(x)cos(x)

y'=x^sin(x)(sin(x)/x + ln(x) cos(x))


Simply a little here:

y'=x^sin(x)(sin(x)/x + ln(x) cos(x))

now I am having problems

x^sin(x) never =0

>> True enough.  But in that case you can ignore the factor, if all you want to do is solve the equation.

and it's undefined at x =0

>> yes, that is also true, and for negative numbers as well.  (negative)^real is undefined.

but since its the same original function its also
undefined so I cant find a critical point

>> Well, it won't be easy, but...

sin(x)/x + ln(x)cos(x)=0

sin(x)/x = -ln(x)cos(x)

tan(x) = -ln(x)x
I don't know enough trigonometry to solve this

...............................
I don't think there is any 'clean' analytic method for finding the solutions.  They definitely exist -- the graph does have extreme points.  But you probably will have to use some numerical method (bisection, Newton, etc.) to find them.

Sorry.

P.S. You can find y' without using logarithmic differentiation, like this:

y= x^(sin(x)) = exp(sin x  ln x)

y' = exp(sin x ln x)(sin x/x + cos x ln x)