Problem 47 - University Calculus: Early Transcendentals

Question

Find the derivatives of the functions in Exercises \(23-50\). $$q=\sin \left(\frac{t}{\sqrt{t+1}}\right)$$

Step-by-Step Solution

Answer

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

Step 1: Identify the Function Structure

The given function is \( q = \sin \left( \frac{t}{\sqrt{t+1}} \right) \). This function is composed of an outer sine function and an inner function \( u(t) = \frac{t}{\sqrt{t+1}} \). To differentiate \( q \), we will need to use the chain rule.

Step 2: Differentiate the Outer Function

Using the chain rule, the derivative of \( q \) with respect to \( t \) is \( \frac{dq}{dt} = \cos(u) \cdot \frac{du}{dt} \) where \( u = \frac{t}{\sqrt{t+1}} \). Thus, we need the derivative of the inner function \( u(t) \).

Step 3: Differentiate the Inner Function

The inner function is \( u(t) = \frac{t}{\sqrt{t+1}} \). We apply the quotient rule: \( \frac{d}{dt} \left( \frac{t}{\sqrt{t+1}} \right) = \frac{\sqrt{t+1} \cdot 1 - t \cdot \frac{1}{2\sqrt{t+1}}}{(\sqrt{t+1})^2} \).

Step 4: Simplify the Derivative of the Inner Function

This becomes \( \frac{(t+1) - \frac{t}{2}}{t+1} = \frac{t + 1 - \frac{t}{2}}{t+1} \). Simplifying further, we get \( \frac{\frac{2t + 2 - t}{2}}{t+1} = \frac{\frac{2 + t}{2}}{t+1} = \frac{t + 2}{2(t+1)} \).

Step 5: Combine Results from Steps 2 and 4

Substitute \( \frac{du}{dt} = \frac{t + 2}{2(t+1)} \) back into the chain rule derivative: \( \frac{dq}{dt} = \cos\left( \frac{t}{\sqrt{t+1}} \right) \cdot \frac{t + 2}{2(t+1)} \).

Step 6: Write the Final Answer

Thus, the derivative of the function \( q = \sin \left( \frac{t}{\sqrt{t+1}} \right) \) with respect to \( t \) is: \[ q'(t) = \cos \left( \frac{t}{\sqrt{t+1}} \right) \cdot \frac{t + 2}{2(t+1)} \]

Key Concepts

Chain Rule

The chain rule is an essential principle in calculus, especially when dealing with composite functions. It's like peeling layers off an onion, except instead of tears, you end up with the derivative of your function! The rule is this: to find the derivative of a composite function, take the derivative of the outer function and multiply it by the derivative of the inner function.

Let's apply this to the function in our exercise, which is structured as an outer sine function and an inner function, which itself includes a fraction. When you hear 'chain rule,' think of these steps:

• Identify the layers: Recognize the outer and inner functions, like how we picked out sine as the outer function here.
• Differentiate: First, take the derivative of the outer function (in this case, the derivative of \( \sin \) is \( \cos \)).
• Multiply: Multiply this by the derivative of the inner function, which requires finding the derivative using the chain or quotient rule as we detail next.

Taking the chain rule makes complex derivatives approachable by breaking them into manageable parts. It's a perfect example of tackling problems layer by layer.

Quotient Rule

When a function is expressed as a quotient, or a fraction of two functions, you'll want to use the quotient rule to find its derivative. Remember the structure of the rule: For a function \( f(t) = \frac{g(t)}{h(t)} \), its derivative is given by:

\[\frac{d}{dt} \left( \frac{g(t)}{h(t)} \right) = \frac{g'(t)h(t) - g(t)h'(t)}{(h(t))^2}\]

The idea here is straightforward yet powerful, just like our function's inner part: \( u(t) = \frac{t}{\sqrt{t+1}} \).

• Numerator Derivative: Take the derivative of the top function, which here is simply 1, since the derivative of \( t \) is 1.
• Denominator Derivative: Take the derivative of the bottom function, which is \( \sqrt{t+1} \). Remembering that \( (\sqrt{t+1})' = \frac{1}{2\sqrt{t+1}} \) can be handy.
• Apply the Rule: With all parts ready, plug them into the quotient rule formula to get your derivative, and simplify where possible.

The quotient rule is indispensable for functions that involve division, allowing you to break down and conquer one of calculus's trickier elements.

Trigonometric Functions

Trigonometric functions like sine and cosine are foundational in calculus, appearing in many natural and engineered systems. When differentiating these functions, it's crucial to know their derivative properties.

For sine and cosine, the following relationships often come in handy:

• The derivative of \( \sin(x) \) is \( \cos(x) \).
• The derivative of \( \cos(x) \) is \( -\sin(x) \).

These rules are based on the trigonometric identities and the circular nature of trigonometric functions.

In the problem we explored, the function \( q = \sin \left( \frac{t}{\sqrt{t+1}} \right) \) necessitated the use of the chain rule because sine operates as the outer function. Thus, knowing these differentiation rules by heart helps apply the chain rule seamlessly.
Make sure to understand how these derivatives interact in the realm of composite functions. It's not just about memorizing equations—it's about interpreting and connecting pieces to form a complete mathematical picture.