Question
Please help me find the derivative of f(x) = x^(2^x)
I got to ln(f(x)) = 2^x(lnx) but now I'm stuck..
I got to ln(f(x)) = 2^x(lnx) but now I'm stuck..
Answers
Reiny
a good start
I would change f(x) to y for easier typing
lny = (2^x)(lnx) , you had that
now use the product rule on the right side
y'/y = (2^x)(1/x) + (lnx)(ln2((2^x)
y' = y( (2^x)/x + ln2(lnx)(2^x))
= x^(2^x)) [ (2^x)/x + ln2(lnx)(2^x) ]
I would change f(x) to y for easier typing
lny = (2^x)(lnx) , you had that
now use the product rule on the right side
y'/y = (2^x)(1/x) + (lnx)(ln2((2^x)
y' = y( (2^x)/x + ln2(lnx)(2^x))
= x^(2^x)) [ (2^x)/x + ln2(lnx)(2^x) ]