Question

(\frac{d}{dt}(ln(3 + 12t)) =)

Explanation:

Step1: Recall the chain rule

The chain rule states that if we have a composite function \( y = \ln(u) \) where \( u = 3 + 12t \), then \( \frac{dy}{dt}=\frac{dy}{du}\cdot\frac{du}{dt} \).
First, find the derivative of the outer function \( y = \ln(u) \) with respect to \( u \): \( \frac{d}{du}(\ln(u))=\frac{1}{u} \).

Step2: Find the derivative of the inner function

Now, find the derivative of the inner function \( u = 3 + 12t \) with respect to \( t \): \( \frac{du}{dt}=\frac{d}{dt}(3 + 12t)=12 \).

Step3: Apply the chain rule

Substitute \( u = 3 + 12t \) and the derivatives we found into the chain rule formula:
\( \frac{d}{dt}(\ln(3 + 12t))=\frac{1}{3 + 12t}\cdot12 \)
Simplify the expression: \( \frac{12}{3 + 12t}=\frac{12}{3(1 + 4t)}=\frac{4}{1 + 4t} \) (or we can leave it as \( \frac{12}{3 + 12t} \), but simplifying gives a neater form).

Answer:

\( \frac{12}{3 + 12t} \) (or simplified as \( \frac{4}{1 + 4t} \))