Sure, let's differentiate the function [tex]\( h(t) = \frac{\ln(9t)}{\ln(72t)} \)[/tex] with respect to [tex]\( t \)[/tex].
### Step-by-Step Solution:
1. Define the function:
[tex]\[
h(t) = \frac{\ln(9t)}{\ln(72t)}
\][/tex]
2. Identify the components:
[tex]\[
u(t) = \ln(9t)
\][/tex]
[tex]\[
v(t) = \ln(72t)
\][/tex]
Therefore,
[tex]\[
h(t) = \frac{u(t)}{v(t)}
\][/tex]
3. Use the quotient rule for differentiation:
The quotient rule states that if [tex]\( h(t) = \frac{u(t)}{v(t)} \)[/tex], then
[tex]\[
h'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{[v(t)]^2}
\][/tex]
4. Find [tex]\( u'(t) \)[/tex]:
[tex]\[
u(t) = \ln(9t)
\][/tex]
Using the chain rule:
[tex]\[
u'(t) = \frac{d}{dt}[\ln(9t)] = \frac{d}{dt}[\ln(9) + \ln(t)] = \frac{d}{dt}[\ln(t)] = \frac{1}{t}
\][/tex]
5. Find [tex]\( v'(t) \)[/tex]:
[tex]\[
v(t) = \ln(72t)
\][/tex]
Similarly, using the chain rule:
[tex]\[
v'(t) = \frac{d}{dt}[\ln(72t)] = \frac{d}{dt}[\ln(72) + \ln(t)] = \frac{d}{dt}[\ln(t)] = \frac{1}{t}
\][/tex]
6. Apply the quotient rule:
[tex]\[
h'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{[v(t)]^2}
\][/tex]
Substituting [tex]\( u'(t) \)[/tex] and [tex]\( v'(t) \)[/tex]:
[tex]\[
u'(t) = \frac{1}{t}, \quad v'(t) = \frac{1}{t}
\][/tex]
Therefore:
[tex]\[
h'(t) = \frac{\left(\frac{1}{t}\right) \ln(72t) - \ln(9t) \left(\frac{1}{t}\right)}{[\ln(72t)]^2}
\][/tex]
7. Simplify the expression:
[tex]\[
h'(t) = \frac{\frac{\ln(72t)}{t} - \frac{\ln(9t)}{t}}{[\ln(72t)]^2}
\][/tex]
Combine the terms in the numerator:
[tex]\[
h'(t) = \frac{\ln(72t) - \ln(9t)}{t [\ln(72t)]^2}
\][/tex]
8. Use the properties of logarithms to simplify further:
[tex]\[
\ln(72t) - \ln(9t) = \ln\left(\frac{72t}{9t}\right) = \ln\left(\frac{72}{9}\right) = \ln(8)
\][/tex]
Hence,
[tex]\[
h'(t) = \frac{\ln(8)}{t [\ln(72t)]^2}
\][/tex]
9. Final expression:
[tex]\[
h'(t) = \frac{\ln(8)}{t [\ln(72t)]^2} + \frac{1}{t \ln(72t)}
\][/tex]
This final expression is equivalent to:
[tex]\[
h'(t) = -\frac{\ln(9t)}{t [\ln(72t)]^2} + \frac{1}{t [\ln(72t)]}
\][/tex]
Thus, the derivative of the function [tex]\( h(t) = \frac{\ln(9t)}{\ln(72t)} \)[/tex] with respect to [tex]\( t \)[/tex] is given by:
[tex]\[
h'(t) = -\frac{\ln(9t)}{t \left( \ln(72t) \right)^2} + \frac{1}{t \ln(72t)}
\][/tex]
