Problem 56 - University Calculus: Early Transcendentals

Question

In Exercises \(51-70,\) find \(d y / d t\). $$y=\left(t^{-3 / 4} \sin t\right)^{4 / 3}$$

Step-by-Step Solution

Answer

\( \frac{dy}{dt} = \frac{4}{3} \left( t^{-3/4} \sin t \right)^{1/3} \left( -\frac{3}{4} t^{-7/4} \sin t + t^{-3/4} \cos t \right) \)

Step 1: Identify the Type of Function

The given function is a composite function where we have a power and a product function combined: \( y = (t^{-3/4} \sin t)^{4/3} \). This suggests using the power rule for derivatives and the chain rule to resolve the composite structure.

Step 2: Apply the Chain Rule

Using the chain rule, if \( y = u^n \), then \( \frac{dy}{dt} = n \cdot u^{n-1} \cdot \frac{du}{dt} \). Here, \( u = t^{-3/4} \sin t \) and \( n = 4/3 \). Substituting, we get: \( \frac{dy}{dt} = \frac{4}{3} (t^{-3/4} \sin t)^{1/3} \cdot \frac{du}{dt} \).

Step 3: Differentiate the Inner Function Using Product Rule

The inner function \( u = t^{-3/4} \sin t \) is a product of two functions. Therefore, we use the product rule: \( \frac{d}{dt}(fg) = f' g + f g' \). For \( f = t^{-3/4} \) and \( g = \sin t \), the derivatives will be \( f' = -\frac{3}{4}t^{-7/4} \) and \( g' = \cos t \). So, \( \frac{du}{dt} = -\frac{3}{4}t^{-7/4} \cdot \sin t + t^{-3/4} \cdot \cos t \).

Step 4: Combine Results From Previous Steps

Substitute \( \frac{du}{dt} \) from Step 3 into the derivative obtained in Step 2: \[ \frac{dy}{dt} = \frac{4}{3} (t^{-3/4} \sin t)^{1/3} \left( -\frac{3}{4}t^{-7/4} \sin t + t^{-3/4} \cos t \right) \].

Step 5: Simplify the Expression

Simplify the expression to complete the differentiation process. The full derivative is represented as a combination of the terms derived. Final form: \[ \frac{dy}{dt} = \frac{4}{3} \left( t^{-3/4} \sin t \right)^{1/3} \left( -\frac{3}{4} t^{-7/4} \sin t + t^{-3/4} \cos t \right) \].

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus used for differentiating composite functions. A composite function is essentially a function within another function, like layers of an onion. For instance, in our original exercise, the expression \((t^{-3/4} \sin t)^{4/3}\) is a composite function. Here, the inner function \(u = t^{-3/4} \sin t\) is "wrapped" in the outer function \(u^{4/3}\). The chain rule helps us peel away these layers by breaking down the differentiation process.

To apply the chain rule, we first differentiate the outer function with respect to the inner one. Then, we multiply that result by the derivative of the inner function with respect to its variable. The formula for the chain rule is as follows:

\[ \frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt} \] where \( y = u^n \) .

In our case, \( u=(t^{-3/4} \sin t)\) and \( n=4/3 \). Differentiating the outer function gives us:

- \( \frac{dy}{du} = \frac{4}{3} u^{1/3} \).

By applying the chain rule, we can find how \( y \) changes with respect to \( t \) by accounting for how the two nested functions interact with each other.

Product Rule

When dealing with a function that is the product of two distinct functions, the product rule becomes essential. It allows us to differentiate expressions like \(t^{-3/4} \sin t\), seen in our example, where the function is split into two parts: \(f(t) = t^{-3/4}\) and \(g(t) = \sin t\).

The product rule states that the derivative of the product of two functions is given by the formula:

\[ \frac{d}{dt}(f(t)g(t)) = f'(t)g(t) + f(t)g'(t) \]

To apply this to our inner function \(u = t^{-3/4} \sin t\), we follow these steps:

• Differentiate \(f(t) = t^{-3/4}\) using the power rule to get \(f'(t) = -\frac{3}{4}t^{-7/4}\).
• Differentiate \(g(t) = \sin t\) to obtain \(g'(t) = \cos t\).
• Substitute the derived values into the product rule formula, resulting in \(-\frac{3}{4}t^{-7/4} \sin t + t^{-3/4} \cos t\).

Applying the product rule here provides the necessary derivative \(\frac{du}{dt}\), which is a key component needed for completing the chain rule differentiation process.

Power Rule

The power rule is one of the most straightforward rules in calculus for finding the derivative of a power function. It is particularly useful when working with functions involving terms like \(t^n\). In our exercise, the power rule is applied both to the inner function \(t^{-3/4}\) and to the composite structure \((t^{-3/4} \sin t)^{4/3}\).

The power rule states that if you have a term \(x^n\), the derivative is:

\[ \frac{d}{dx}(x^n) = n \cdot x^{n-1} \]

Using the power rule, we differentiate the inner term \(t^{-3/4}\) as follows:

- \( \frac{d}{dt}(t^{-3/4}) = -\frac{3}{4} \cdot t^{-7/4}\).

For the outer expression, \((u)^{4/3}\), we apply the power rule in the context of the chain rule. This involves taking the exponent \(\frac{4}{3}\), multiplying it by the function raised to one less power, and then multiplying the result by the derivative of the inner function.

The combination of the chain rule and power rule allows us to effectively handle and decrypt the layers present in composite functions like our original problem. By understanding these underlying rules, solving such derivatives becomes more manageable and intuitive.