Problem 62 - College Algebra and Calculus: An Applied Approach

Question

Find the derivative of the function. State which differentiation rule(s) you used to find the derivative. $$ y=t \sqrt{t+1} $$

Step-by-Step Solution

Answer

The derivative of \(y = t \sqrt{t+1}\) is \(y' = \sqrt{t+1} + \frac{t}{2\sqrt{t+1}}\) using the product and chain rules of differentiation.

Step 1: Identify the Functions

First, identify the two functions involved in the equation, which will be differentiated using the product rule. The functions are \(f(t)=t\) and \(g(t)=\sqrt{t+1}\).

Step 2: Find the Derivatives

Next, find the derivatives of the two functions. Remember, the power rule states that the derivative of \(t^n\) is \(nt^{n-1}\), and the chain rule states that the derivative of \(y = f(g(t))\) is \(y' = f'(g(t)) \cdot g'(t)\). \n\nThe derivative of \(f(t)=t\) is \(f'(t)=1\). The derivative of \(g(t)=\sqrt{t+1}\) requires the chain rule since it has a function within a function. Substituting \(u = t + 1\), the function becomes \(g(u)=\sqrt{u}\), which is \(u^{\frac{1}{2}}\). Using the power rule, the derivative is \(\frac{1}{2}u^{-\frac{1}{2}}\). But we substituted \(u = t + 1\), so the derivative becomes \(\frac{1}{2}(t+1)^{-\frac{1}{2}}\), multiplied by the derivative of \(u = t + 1\), which is 1. Hence, \(g'(t)=\frac{1}{2}(t+1)^{-\frac{1}{2}}\).

Step 3: Apply the Product Rule

Now, apply the product rule, \( (f \cdot g)' = f' \cdot g + f \cdot g'\). Substituting the functions and their derivatives into this rule gives \(y' = 1 \cdot \sqrt{t+1} + t \cdot \frac{1}{2}(t+1)^{-\frac{1}{2}} = \sqrt{t+1} + \frac{t}{2\sqrt{t+1}}\).