Problem 58 - University Calculus: Early Transcendentals

Question

In Exercises \(51-70,\) find \(d y / d t\). $$y=\left(e^{\sin (t / 2)}\right)^{3}$$

Step-by-Step Solution

Answer

The derivative is \( \frac{3}{2} e^{3\sin(t/2)} \cos(t/2) \).

Step 1: Identify the Function and Rewrite

The given function is \( y = \left( e^{\sin(t/2)} \right)^3 \). Recognize this as a composition of functions: an outer function \( u^3 \) and an inner function \( u = e^{\sin(t/2)} \). To simplify differentiation, rewrite this as \( y = \left( e^{u} \right)^3 = u^3 \), where \( u = e^{\sin(t/2)} \).

Step 2: Differentiate the Outer Function

Use the power rule to differentiate the outer function \( u^3 \) with respect to \( u \):\[ \frac{d}{du}(u^3) = 3u^2. \]

Step 3: Differentiate the Inner Function

To differentiate \( u = e^{\sin(t/2)} \) with respect to \( t \), apply the chain rule: - The derivative of \( e^{\sin(t/2)} \) with respect to \( \sin(t/2) \) is \( e^{\sin(t/2)} \). - The derivative of \( \sin(t/2) \) with respect to \( t \) is \( \cos(t/2) \cdot \frac{1}{2} \). Therefore, \[ \frac{du}{dt} = e^{\sin(t/2)} \cdot \cos(t/2) \cdot \frac{1}{2}. \]

Step 4: Apply the Chain Rule

Using the chain rule, combine the derivatives of the outer and the inner functions to find \( \frac{dy}{dt} \):\[ \frac{dy}{dt} = \frac{dy}{du} \times \frac{du}{dt}. \]Substitute the expressions derived earlier:\[ \frac{dy}{dt} = 3u^2 \cdot e^{\sin(t/2)} \cdot \cos(t/2) \cdot \frac{1}{2}. \]Since \( u = e^{\sin(t/2)} \), substitute back:\[ \frac{dy}{dt} = 3(e^{\sin(t/2)})^2 \cdot e^{\sin(t/2)} \cdot \frac{1}{2} \cdot \cos(t/2). \]

Key Concepts

Chain Rule

The chain rule is a fundamental concept in differential calculus. It is used when dealing with composite functions, where one function is nested inside another. Imagine peeling layers of an onion; this is similar to how we differentiate step by step using the chain rule.
To apply the chain rule, let's say we have a function defined as \(y = f(g(t))\). This means that \(y\) depends on \(g(t)\), and \(g(t)\) depends on \(t\). The chain rule helps us find the derivative of \(y\) with respect to \(t\). The formula for the chain rule is as follows:

• \(\frac{dy}{dt} = \frac{dy}{dg} \times \frac{dg}{dt}\)

Using separate differentiation steps makes dealing with complex functions much more manageable. You differentiate the outer function while treating the inner function as a variable. Then, differentiate the inner function with respect to the variable \(t\), and combine the results.
In the original exercise, the function \(y = (e^{\sin(t/2)})^3\) involves an exponential function \(u = e^{\sin(t/2)}\) nested inside a power function \(u^3\). Using the chain rule helps to break this complex differentiation task into smaller, manageable pieces.

Derivative of Exponential Functions

Exponential functions, characterized by expressions of the form \(e^x\), have a unique property: they have the same derivative as the function itself. This makes working with them particularly elegant in calculus.
For an exponential function like \(e^{\sin(t/2)}\), when finding its derivative, we treat the exponent as its own function, \(v(t)=\sin(t/2)\). The derivative of \(e^v\) with respect to \(v\) remains \(e^v\), because the derivative of the exponential function \(e^x\) is itself:

• \(\frac{d}{dv}(e^v) = e^v\)

After finding the derivative, there's the task of applying the chain rule. Thus one must consider how the exponent itself changes with respect to \(t\). In this case, the derivative of \(\sin(t/2)\) with respect to \(t\) needs to be calculated using basic trigonometric derivatives.
Remember that differentiation of this nature demands careful observation of how each composite part of the function affects the whole.

Power Rule for Derivatives

The power rule is one of the most straightforward and frequently used rules in differential calculus. It's used to find the derivative of a function of the form \(x^n\), where \(n\) is any real number. The rule states that if you have a function \(y = x^n\), its derivative with respect to \(x\) is:

• \(\frac{d}{dx}(x^n) = nx^{n-1}\)

This rule simplifies finding the derivative of polynomial and other power-based functions. In the case of the original exercise, the power rule was applied to the outer function, \(u^3\), which resulted in finding its derivative as \(3u^2\).
Power functions form the backbone of many calculus problems, making it essential to understand how to apply the power rule quickly and accurately. By understanding how to apply the power rule, one can navigate through more complex differentiation tasks that involve power terms alongside other functions.
The power rule works seamlessly with the chain rule, as seen in the exercise, by allowing the differentiation of nested functions, stepping one level deeper each time.