Problem 60 - University Calculus: Early Transcendentals

Question

In Exercises \(51-70,\) find \(d y / d t\). $$y=\left(\frac{3 t-4}{5 t+2}\right)^{-5}$$

Step-by-Step Solution

Answer

\( \frac{dy}{dt} = -130 \cdot \left( \frac{3t - 4}{5t + 2} \right)^{-6} \cdot (5t + 2)^{-2} \).

Step 1: Identify the Function Structure

The given function is \(y = \left( \frac{3t - 4}{5t + 2} \right)^{-5}\). This follows the general form of \((u(t))^n\) where \(u(t) = \frac{3t - 4}{5t + 2}\) and \(n = -5\). We will apply the chain rule to differentiate this function.

Step 2: Apply the Chain Rule

The chain rule states that \( \frac{dy}{dt} = n \cdot (u(t))^{n-1} \cdot \frac{du}{dt} \). Thus, we set up the derivative as follows:\( \frac{dy}{dt} = -5 \cdot \left( \frac{3t - 4}{5t + 2} \right)^{-6} \cdot \frac{d}{dt}\left(\frac{3t - 4}{5t + 2}\right) \). Now, we need to find \( \frac{d}{dt}(u(t)) \).

Step 3: Differentiate the Inner Function

The function \(u(t) = \frac{3t - 4}{5t + 2}\) requires the quotient rule for differentiation, which states \( \frac{d}{dt}\left(\frac{f(t)}{g(t)}\right) = \frac{f'(t)g(t) - f(t)g'(t)}{(g(t))^2} \).Here, \(f(t) = 3t - 4\) and \(g(t) = 5t + 2\). Differentiate both: \(f'(t) = 3\) and \(g'(t) = 5\).

Step 4: Apply the Quotient Rule

Substitute the derivatives into the quotient rule formula:\[ \frac{du}{dt} = \frac{(3)(5t + 2) - (3t - 4)(5)}{(5t + 2)^2} \]Simplify the numerator: \(3(5t + 2) - 5(3t - 4) = 15t + 6 - 15t + 20 = 26\).Thus, \( \frac{du}{dt} = \frac{26}{(5t + 2)^2} \).

Step 5: Substitute Back into the Chain Rule

Now, substitute \( \frac{du}{dt} = \frac{26}{(5t + 2)^2} \) back into the chain rule expression:\[ \frac{dy}{dt} = -5 \cdot \left( \frac{3t - 4}{5t + 2} \right)^{-6} \cdot \frac{26}{(5t + 2)^2} \].

Step 6: Simplify the Expression

Combine and simplify the expression:\[ \frac{dy}{dt} = -130 \cdot \frac{1}{(5t + 2)^2} \cdot \left( \frac{3t - 4}{5t + 2} \right)^{-6} \]Final simplified form:\[ \frac{dy}{dt} = -130 \cdot \left( \frac{3t - 4}{5t + 2} \right)^{-6} \cdot (5t + 2)^{-2} \].

Key Concepts

Chain Rule

The chain rule is an essential tool in calculus, especially for differentiating complex functions. It helps us find the derivative of composite functions, where one function is composed inside another. Imagine a function defined as \( y = (u(t))^n \) with \( u(t) \) as a function of \( t \). The chain rule tells us how to differentiate such a function efficiently.

The formula is given by:

• \( \frac{dy}{dt} = n \cdot (u(t))^{n-1} \cdot \frac{du}{dt} \)

In this formula, \( n \) is the power applied to \( u(t) \), \( (u(t))^{n-1} \) is the new exponent after differentiation, and \( \frac{du}{dt} \) is the derivative of the inner function \( u(t) \).

For example, with \( y = \left( \frac{3t - 4}{5t + 2} \right)^{-5} \), the outer function is a power function, and the inner function \( u(t) = \frac{3t - 4}{5t + 2} \). Differentiating requires applying the chain rule efficiently by finding the derivative of the inner function separately before substituting back into the chain rule formula.

Quotient Rule

The quotient rule is a technique for finding the derivative of a fraction of two functions. When you have a function like \( u(t) = \frac{f(t)}{g(t)} \), the quotient rule is the way to go. This rule helps deal with fractions where both the numerator and the denominator are differentiable functions.

The derivative formula for the quotient rule is:

• \( \frac{d}{dt}\left(\frac{f(t)}{g(t)}\right) = \frac{f'(t)g(t) - f(t)g'(t)}{(g(t))^2} \)

Here, \( f(t) \) and \( g(t) \) are functions of \( t \). The terms \( f'(t) \) and \( g'(t) \) are their respective derivatives.

In our example, \( f(t) = 3t - 4 \) and \( g(t) = 5t + 2 \). Their derivatives are \( f'(t) = 3 \) and \( g'(t) = 5 \). Applying the quotient rule gives us \( \frac{du}{dt} = \frac{26}{(5t + 2)^2} \), which we then use in the chain rule.

Derivatives

Derivatives are the backbone of calculus, providing a way to understand how functions change. They tell us the rate at which a function is changing at any given point — a type of 'instantaneous speed' of a function.

A derivative of a function \( f(t) \) with respect to \( t \) is represented by \( \frac{df}{dt} \). This notation indicates the rate of change of \( f \) expressed per unit change in \( t \).

For the function \( y = \left( \frac{3t - 4}{5t + 2} \right)^{-5} \), finding the derivative involves both the chain rule and the quotient rule. By applying these rules, the derivative is derived:

• The chain rule handles the outer function exponent conversion.
• The quotient rule differentiates the inner rational function.

Finally, the combination of these derivatives is simplified to find \( \frac{dy}{dt} \), fully describing how \( y \) changes with \( t \). It's this mastery over derivatives that gives calculus its power to solve real-world problems.