Problem 59 - University Calculus: Early Transcendentals

Question

In Exercises \(51-70,\) find \(d y / d t\). $$y=\left(\frac{t^{2}}{t^{3}-4 t}\right)^{3}$$

Step-by-Step Solution

Answer

The derivative is \(-3 \cdot \frac{t^4(t^2 + 4)}{(t^3 - 4t)^4}\)."

Step 1: Identify the function.

The function given is a composite function, which appears as a rational function raised to the power of 3. Specifically, it is written as \( y = \left(\frac{t^{2}}{t^{3}-4t}\right)^{3} \).

Step 2: Apply the chain rule.

The chain rule of differentiation states \( \frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt} \) where \( u = \left(\frac{t^{2}}{t^{3} - 4t}\right) \). Differentiating \( y = u^3 \) gives \( \frac{dy}{du} = 3u^2 \).

Step 3: Differentiate the inner function.

Differentiate \( u = \frac{t^{2}}{t^{3}-4t} \) using the quotient rule. The quotient rule is \( \left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2} \). Here, let \( f(t) = t^2 \) and \( g(t) = t^3 - 4t \).

Step 4: Apply the quotient rule.

Find \( f' = 2t \) and \( g' = 3t^2 - 4 \). Substitute into the quotient rule formula:\[\frac{du}{dt} = \frac{(2t)(t^3 - 4t) - (t^2)(3t^2 - 4)}{(t^3 - 4t)^2} \]Simplifying, \((2t)(t^3 - 4t) = 2t^4 - 8t^2\) and \((t^2)(3t^2 - 4) = 3t^4 - 4t^2\), so:\[\frac{du}{dt} = \frac{2t^4 - 8t^2 - 3t^4 + 4t^2}{(t^3 - 4t)^2} = \frac{-t^4 - 4t^2}{(t^3 - 4t)^2}\]which simplifies further to\[\frac{du}{dt} = \frac{-t^2(t^2 + 4)}{(t^3 - 4t)^2}\].

Step 5: Combine using the chain rule.

Substitute \( u \) and \( \frac{du}{dt} \) back into the chain rule:\[\frac{dy}{dt} = 3u^2 \cdot \frac{-t^2(t^2 + 4)}{(t^3 - 4t)^2}\]Since \( u = \frac{t^2}{t^3 - 4t} \), \( u^2 = \left(\frac{t^2}{t^3 - 4t}\right)^2 \), hence:\[\frac{dy}{dt} = 3 \cdot \left(\frac{t^2}{t^3 - 4t}\right)^2 \cdot \frac{-t^2(t^2 + 4)}{(t^3 - 4t)^2}\].

Step 6: Simplify the derivative expression.

Simplify the expression \[3 \cdot \left(\frac{t^2}{t^3 - 4t}\right)^2 \cdot \frac{-t^2(t^2 + 4)}{(t^3 - 4t)^2}\]:\[\frac{dy}{dt} = 3 \cdot \frac{t^4(t^2 + 4)}{(t^3 - 4t)^4}\]Rewriting using the negative sign:\[\frac{dy}{dt} = -3 \cdot \frac{t^4(t^2 + 4)}{(t^3 - 4t)^4}\].

Key Concepts

Chain Rule

Understanding the chain rule is essential when dealing with composite functions. The chain rule is a fundamental technique in calculus for finding the derivative of composite functions. Imagine you have a situation where one function is "inside" another. We typically express this as \( y = f(g(x)) \), where \( g(x) \) is the inside function and \( f \) is the outside function. The chain rule states that the derivative of the composite function is the derivative of the outside function evaluated at the inside function times the derivative of the inside function. This can be expressed as:

• \( \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \)

In our exercise, we applied the chain rule by first identifying \( u = \frac{t^2}{t^3 - 4t} \) as the inside function in the composite \( y = u^3 \). The chain rule was used to differentiate \( y \) by finding the derivative of \( u^3 \) and then multiplying it by the derivative of \( u \). This is what helped link the change in \( y \) directly to the change in \( t \). The chain rule is powerful because it simplifies the differentiation of complex functions into manageable steps.

Quotient Rule

The quotient rule is indispensable when differentiating functions expressed as fractions, or more formally, the quotient of two functions. If you have a function \( \frac{f(x)}{g(x)} \), the quotient rule states:

• \( \left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2} \)

This rule essentially tells us to take the derivative of the top function \( f(x) \) times the bottom function \( g(x) \) minus the top function times the derivative of the bottom function, all over the bottom function squared. In our example, we determined \( f(t) = t^2 \) and \( g(t) = t^3 - 4t \). By applying the quotient rule, we calculated \( \frac{du}{dt} \) which was necessary to use with the chain rule later on. The process involved careful substitution and simplification to ensure that each component was accurately differentiated, maintaining the integrity of the original function. As such, the quotient rule assists in breaking down and managing more complex rational expressions.

Composite Functions

Composite functions consist of an arrangement where one function's output becomes another function's input. These are encountered often in calculus problems where multiple operations are layered on each other. For any composite expression \( y = f(g(x)) \), \( g(x) \) feeds into \( f \), creating a layered effect. In this exercise, the inner function \( u = \frac{t^2}{t^3 - 4t} \) is embedded within another function that cubes its result. This presents \( y = u^3 \) as a composite function form.
To solve for derivatives in composite functions, it is crucial to deconstruct them into their constituent functions. The chain rule becomes your guide to perform differentiation as each layer is dependent on the previous. By acknowledging and extracting the inner function first, you can systematically approach the derivative problem. For our example, recognizing \( u \) as a composite ingredient made it feasible to apply both the quotient and chain rules sequentially, simplifying the differentiating process. Composite functions might seem challenging, but with the right breakdown, they become manageable and clear.