Using logarithmic diffrentiation find dy/dx of y=10^x^x?

Question

I want steps now itself urgently for above question For crosscheacking answer is Dy/dx=10^x^x*log10(1+logx)

Answer 1

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.