Question

If \(s = \sqrt{t + 1}, x = \log s\) and \(y = 6x + 3\), then \(\frac{dy}{dt} =\)

• A. \(\frac{2}{\sqrt{t + 1}}\)
• B. \(\frac{6}{t + 1}\)
• C. \(3\sqrt{t + 1}\)
• D. \(\frac{3}{t + 1}\)
• E. \(\frac{3}{\sqrt{t + 1}}\)

Hint:

Always simplify logarithmic expressions before differentiating. Using \(\log(\sqrt{u}) = \frac{1}{2}\log u\) turns a messy chain rule problem into a simple single-step derivative.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This is a composite function derivative. We can use the chain rule: \(\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{ds} \cdot \frac{ds}{dt}\).
Alternatively, express \(y\) directly in terms of \(t\).

Step 2: Key Formula or Approach:
1. \(y = 6 \log(\sqrt{t+1}) + 3\).
2. Property of logs: \(\log(a^b) = b \log a\).

Step 3: Detailed Explanation:
Substitute \(s\) and \(x\) into \(y\):
\[ x = \log(t + 1)^{1/2} = \frac{1}{2} \log(t + 1) \]
Substitute \(x\) into \(y\):
\[ y = 6 \cdot \left( \frac{1}{2} \log(t + 1) \right) + 3 \]
\[ y = 3 \log(t + 1) + 3 \]
Differentiate with respect to \(t\):
\[ \frac{dy}{dt} = 3 \cdot \frac{d}{dt}[\log(t + 1)] + 0 \]
\[ \frac{dy}{dt} = 3 \cdot \frac{1}{t + 1} = \frac{3}{t + 1} \]

Step 4: Final Answer:
The derivative is \(\frac{3}{t + 1}\).