# Just a simple derivation question.

• April 28th 2008, 07:36 PM
Jeavus
Just a simple derivation question.
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?
• April 28th 2008, 07:41 PM
Mathstud28
Quote:

Originally Posted by Jeavus
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)

Solving for y' we get $y'=\frac{1}{ln(2)ln(3)3^yt}$

(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?

here is whta I would do $y=log_3(log_2(t))$

so then $3^y=log_2(t)$

Differenatiating both sides we would get $\ln(3)3^{y}y'=\frac{1}{ln(2)t}$

and since $y=\log_3(\log_2(x))$

we have $y'=\frac{1}{\ln(2)\ln(3)t\log_2(t)}$

Evaluating we get $y'(8)=\frac{1}{\ln(2)\ln(3)\cdot{8}\cdot\log_2(2^3 )}=\frac{1}{\ln(2)\ln(3)\cdot{24}}$

which is what you asked for