Verify the relation f'(x)/f(x)=1+lnx, if f(x)=e^(ln((x)^x))?
2 Answers
justaguide | College Teacher | (Level 2) Distinguished Educator
The function f(x) = e^(ln((x)^x))
taking the log to the base e of both the sides we get:
ln f(x) = ln x^x
differentiate both the sides
f'(x)/f(x) = [ln x^x]'
=> f'(x)/f(x) = [x*ln x]'
=> f'(x)/ f(x) = x*(1/x) + ln x
=> f'(x) / f(x) = 1 + ln x
This proves that f'(x)/f(x) = 1 + ln x
giorgiana1976 | College Teacher | (Level 3) Valedictorian
To verify the given identity, we'll have to differentiate the function f(x):
f'(x) =e^[ln((x)^x)]*[ln((x)^x)]'
[ln((x)^x)]' = [x*ln(x)]'
We'll use the product rule:
(u*v)' = u'*v + u*v'
[x*ln(x)]' = x'*ln x + x*[ln(x)]'
[x*ln(x)]' = ln x + x/x
[x*ln(x)]' = ln x + 1
So, the derivative of the function f(x) is:
f'(x) = e^[ln((x)^x)]*(ln x + 1)
But e^[ln((x)^x)] = f(x).
f'(x) = f(x)*(ln x + 1)
We'll divide both sides by f(x) and we'll get:
f'(x)/ f(x) = (ln x + 1) q.e.d.
We notice that the identity f'(x)/f(x)=1+lnx is verified.