Question: Using logarithmic diffrentiation find dy/dx of y=10^x^x?
y=10^x^x
There are two ways this could be interpreted. As y = (10^x)^x or as y = 10^(x^x). Brackets help :)
Use y = 10^(x^x).
Take logs of both sides.
ln(y) = (x^x).ln(10)
Differentiating both sides wrt x.
(1/y).y' = X'.ln(10), where X = x^x ------------------- (1)
Taking logs of X = x^x,
ln(X) = x.ln(x)
Differentiating both sides wrt x.
(1/X).X' = ln(x) + 1
X' = (x^x)(1 + ln(x))
Substitute for X' into (1)
(1/y).y' = (x^x)(1 + ln(x)).ln(10),
y' = 10^(x^x).(x^x)ln(10)(1 + ln(x))
I think that the answer you gave, 10^x^x*log10(1+logx), has a factor of (x^x) missing.