Question

find the derivative of the function f(t), below.
f(t) = ln(ln(2t)) + ln(ln 8)
f(t) =

Explanation:

Step1: Recall derivative rules

The derivative of a sum of functions is the sum of their derivatives. Also, the derivative of a constant is 0. Since $\ln(\ln 8)$ is a constant, its derivative is 0. We only need to find the derivative of $y = \ln(\ln(2t))$.

Step2: Apply chain - rule

Let $u=\ln(2t)$ and $y = \ln(u)$. The chain - rule states that $\frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt}$. First, find $\frac{dy}{du}$. The derivative of $y=\ln(u)$ with respect to $u$ is $\frac{1}{u}$.

Step3: Find $\frac{du}{dt}$

Now, find the derivative of $u = \ln(2t)$ with respect to $t$. Using the chain - rule again (or the fact that $\ln(2t)=\ln 2+\ln t$ and the derivative of $\ln 2$ is 0 and the derivative of $\ln t$ is $\frac{1}{t}$), we have $\frac{du}{dt}=\frac{1}{t}$.

Step4: Calculate $\frac{dy}{dt}$

Substitute $u = \ln(2t)$ back into $\frac{dy}{du}$ and multiply by $\frac{du}{dt}$. We get $\frac{dy}{dt}=\frac{1}{u}\cdot\frac{du}{dt}=\frac{1}{\ln(2t)}\cdot\frac{1}{t}=\frac{1}{t\ln(2t)}$.

Answer:

$\frac{1}{t\ln(2t)}$