h(t) = log(base3)[log(base2)(t)]
Evaluate for h'(8).
This is what I have so far:
Let g(x) = log(base2)(t)
g'(x) = 1 / (t ln2)
Therefore.
h'(t) = 1 / (t ln2) / (log(base 2)t)(ln3)
h'(t) = 1 / (t)(ln2)(log(base2)t)(ln3)
(using the rule: log(base x)(g(x)) = g'(x) / g(x)ln(b)
The answer in the back of my textbook gives:
h'(8) = 1 / 24(ln2)(ln3)
How do I go from where I am to where the book is?