Problem 37 - Thomas Calculus

Question

Find the derivatives of the functions in Exercises \(19-38\) $$ q=\sin \left(\frac{t}{\sqrt{t+1}}\right) $$

Step-by-Step Solution

Answer

\( q'(t) = \cos\left(\frac{t}{\sqrt{t+1}}\right) \cdot \frac{t+2}{2(t+1)^{3/2}} \).

Step 1: Identify the Function and Derivative Needed

We need to find the derivative of the function \( q(t) = \sin\left(\frac{t}{\sqrt{t+1}}\right) \). This requires using both the chain rule and the quotient rule.

Step 2: Apply the Chain Rule

First, identify the outer and inner functions. The function \( q(t) = \sin(u) \) where \( u = \frac{t}{\sqrt{t+1}} \). Differentiate the outer function: the derivative of \( \sin(u) \) is \( \cos(u) \), so \( \frac{dq}{du} = \cos(u) \).

Step 3: Differentiate the Inner Function Using the Quotient Rule

Next, we apply the quotient rule to find the derivative of \( u = \frac{t}{\sqrt{t+1}} \). The quotient rule states: \[\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}\]For \( f(t) = t \) and \( g(t) = \sqrt{t+1} \). Their derivatives are \( f'(t) = 1 \) and \( g'(t) = \frac{1}{2\sqrt{t+1}} \).

Step 4: Substitute into the Quotient Rule Formula

Apply the derivatives into the quotient rule formula:\[\frac{du}{dt} = \frac{1 \cdot \sqrt{t+1} - t \cdot \frac{1}{2\sqrt{t+1}}}{(\sqrt{t+1})^2}\]Simplify the expression to obtain:\[\frac{du}{dt} = \frac{\sqrt{t+1} - \frac{t}{2\sqrt{t+1}}}{t+1}\]Further simplify:\[\frac{du}{dt} = \frac{2(t+1) - t}{2\sqrt{t+1}(t+1)} = \frac{2(t+1) - t}{2(t+1)^{3/2}}\] Simplify to get \[ \frac{du}{dt} = \frac{t+2}{2(t+1)^{3/2}} \].

Step 5: Combine Chain Rule and Quotient Rule Results

Substitute \( \frac{du}{dt} \) back into the chain rule:\[\frac{dq}{dt} = \cos\left(\frac{t}{\sqrt{t+1}}\right) \cdot \frac{t+2}{2(t+1)^{3/2}}\]

Step 6: Final Expression for the Derivative

The final derivative of the function \( q(t) = \sin\left(\frac{t}{\sqrt{t+1}}\right) \) is:\[q'(t) = \cos\left(\frac{t}{\sqrt{t+1}}\right) \cdot \frac{t+2}{2(t+1)^{3/2}}\]

Key Concepts

Chain Rule

The chain rule is an essential technique in calculus used to find the derivative of compositions of functions. Imagine you're peeling layers of an onion; think of the outer layer as one function and each inner layer as another function. In mathematical terms, if you have a function like \( q(t) = \sin(u) \), where \( u \) is itself another function of \( t \), you'll need to differentiate each layer step by step.
Here's how it works in this example:

• Start by identifying the outer function, which is \( \sin(u) \), and note that its derivative is \( \cos(u) \).
• The inner function here is \( u = \frac{t}{\sqrt{t+1}} \).
  
• Differentiate the outer function, applying \( \frac{dq}{du} = \cos(u) \).
• Don't forget that you also need to differentiate the inner function, which is where the quotient rule comes in.

By employing the chain rule, one effectively multiplies the derivative of the outer function by the derivative of the inner function, which we will explore in detail next. This process ensures that the rate of change of the overall function is properly captured.

Quotient Rule

The quotient rule is your go-to tool when dealing with the derivative of a ratio. Anytime a problem requires differentiating a function like \( \frac{f(t)}{g(t)} \), the quotient rule comes into play. The formula for this rule is:\[\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}.\]In our exercise, \( u(t) = \frac{t}{\sqrt{t+1}} \) fits this scenario perfectly.
Here is a step-by-step breakdown:

• Identify \( f(t) = t \) and \( g(t) = \sqrt{t+1} \).
• The derivatives of these are \( f'(t) = 1 \) and \( g'(t) = \frac{1}{2\sqrt{t+1}} \).
• Plug these derivatives into the quotient rule formula: \[\frac{du}{dt} = \frac{1 \cdot \sqrt{t+1} - t \cdot \frac{1}{2\sqrt{t+1}}}{(\sqrt{t+1})^2}\]
• Simplify the expression to achieve: \[\frac{t+2}{2(t+1)^{3/2}}.\]

The quotient rule allows us to manage complex derivatives by structuring them into simpler, manageable steps. It's particularly useful when combined with the chain rule to handle intricate layers of functions.

Trigonometric Functions

Trigonometric functions are omnipresent in calculus, especially when addressing oscillatory behaviors or periodic phenomena. In our scenario, we're dealing with the sine function, \( q(t) = \sin\left(\frac{t}{\sqrt{t+1}}\right) \).
Here's what you need to know about differentiating these functions:

• The derivative of \( \sin(u) \) is \( \cos(u) \).
• Once the inner derivative is calculated using the quotient rule, you multiply it by \( \cos(u) \) as dictated by the chain rule.
• Trigonometric derivatives often lead to additional simplifications, like identities, which can streamline calculations.

By combining the knowledge of both the chain rule and the quotient rule, you effectively tackle the derivative involving the sine function. Trigonometric functions not only introduce unique derivatives but also bring in their identities, enhancing the problem-solving toolkit at your disposal. This multilayer application showcases the interconnectedness of calculus rules to solve multifaceted problems easily.