Question

find the derivative of the function.\\( g(t) = \frac{16 \log_{4}(t)}{t} \\)\\( g(t) = \frac{16(1 - t\log(4)(t))}{t^{2}\ln(4)} \\) \\(\times\\)\
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Explanation:

Step1: Identify quotient rule form

Let $f(t)=16\log_4(t)$, $h(t)=t$, so $g(t)=\frac{f(t)}{h(t)}$. Quotient rule: $g'(t)=\frac{f'(t)h(t)-f(t)h'(t)}{[h(t)]^2}$

Step2: Derive $f'(t)$ using log rule

$\log_4(t)=\frac{\ln(t)}{\ln(4)}$, so $f(t)=\frac{16\ln(t)}{\ln(4)}$. Then $f'(t)=\frac{16}{t\ln(4)}$

Step3: Derive $h'(t)$

$h(t)=t$, so $h'(t)=1$

Step4: Substitute into quotient rule

$$\begin{align*} g'(t)&=\frac{\frac{16}{t\ln(4)} \cdot t - 16\log_4(t) \cdot 1}{t^2}\\ &=\frac{\frac{16}{\ln(4)} - 16\cdot\frac{\ln(t)}{\ln(4)}}{t^2} \end{align*}$$

Step5: Simplify the expression

Factor out $\frac{16}{\ln(4)}$:

$$ g'(t)=\frac{16(1 - \ln(t))}{t^2\ln(4)} $$

Answer:

$\boldsymbol{\frac{16(1 - \ln(t))}{t^2\ln(4)}}$