Question

Find the indicated derivative. $$ D_{t}\left(\frac{3 t-2}{t+5}\right)^{3} $$

Short answer

The derivative is \( \frac{51(3t-2)^2}{(t+5)^4} \).

Step-by-step solution

Understand the Function and Differentiation

The function given is \( \left(\frac{3t-2}{t+5}\right)^3 \), and we need to find its derivative with respect to \( t \). We will apply the chain rule and the quotient rule to solve this.

Apply the Chain Rule

Set \( u = \frac{3t-2}{t+5} \). Then the original function becomes \( u^3 \). By the chain rule, the derivative of \( u^3 \) with respect to \( t \) is \( 3u^2 \cdot \frac{du}{dt} \). So, we first need to find \( \frac{du}{dt} \).

Apply the Quotient Rule to Find \( \frac{du}{dt} \)

To find \( \frac{du}{dt} \) where \( u = \frac{3t-2}{t+5} \), use the quotient rule: \( \frac{d}{dt}\left(\frac{f}{g}\right) = \frac{g \cdot f' - f \cdot g'}{g^2} \). Here, \( f = 3t-2 \) and \( g = t+5 \). Thus, \( f' = 3 \) and \( g' = 1 \).

Differentiate Using the Quotient Rule

Substitute to get \( \frac{du}{dt} = \frac{(t+5)(3) - (3t-2)(1)}{(t+5)^2} \). Simplify this to get \( \frac{du}{dt} = \frac{3t + 15 - 3t + 2}{(t+5)^2} = \frac{17}{(t+5)^2} \).

Substitute Back into the Chain Rule Expression

Now using the chain rule expression \( 3u^2 \cdot \frac{du}{dt} \), substitute \( u = \frac{3t-2}{t+5} \) and \( \frac{du}{dt} = \frac{17}{(t+5)^2} \). This gives us \( 3\left(\frac{3t-2}{t+5}\right)^2 \cdot \frac{17}{(t+5)^2} \).

Simplify the Derivative Expression

Simplify the expression from the previous step to get \( \frac{51(3t-2)^2}{(t+5)^4} \). This is the simplified form of the derivative.

Key concepts

Chain Rule

The Chain Rule is a critical strategy in calculus for finding the derivative of composite functions. If you have a function that is composed of two or more functions, then the Chain Rule helps break down the differentiation into manageable parts.
In simpler terms, suppose you have a function like \( (g(t))^n \), where \( g(t) \) itself is a function of \( t \). The Chain Rule states that the derivative of \( u^n \) with respect to \( t \) is given by \[ n \, u^{n-1} \cdot \frac{du}{dt} \] where \( u = g(t) \).

• Identify the inner function as \( u \).
• Determine the derivative of the outer function with respect to \( u \).
• Multiply it by the derivative of the inner function with respect to \( t \), which is \( \frac{du}{dt} \).

When solving the problem \( \left( \frac{3t-2}{t+5} \right)^3 \), the Chain Rule simplifies the process by setting \( u = \frac{3t-2}{t+5} \), allowing for a seamless calculation by handling the exponent \( 3 \) separately from the fraction expression.

Quotient Rule

The Quotient Rule is an essential differentiation technique for dealing with divisions of functions. If you need to differentiate a function defined as the quotient of two differentiable functions, such as \( \frac{f(t)}{g(t)} \), the Quotient Rule provides the formula: \[ \frac{d}{dt} \left( \frac{f}{g} \right) = \frac{g \cdot f' - f \cdot g'}{g^2} \]

• Identify the numerator as \( f \) and the denominator as \( g \).
• Determine their respective derivatives, \( f' \) and \( g' \).
• Apply the formula to get the rate of change of the whole fraction.

In the given exercise, use the Quotient Rule on \( u = \frac{3t-2}{t+5} \). Here, \( f = 3t-2 \) with \( f' = 3 \) and \( g = t+5 \) with \( g' = 1 \). This simplifies to \( \frac{17}{(t+5)^2} \), an important component needed for the subsequent Chain Rule application.

Differentiation Techniques

Differentiation Techniques in calculus provide methods to find the rate of change of functions. They are fundamental to understanding how functions behave and change.Several key techniques include the Chain Rule and the Quotient Rule, which are tackled above.

These techniques are used to handle more complex functions where one simple method isn't sufficient. Differentiating powers, products, and quotients might require combining these strategies to resolve more intricate problems. For example, analyzing the term \( \left( \frac{3t-2}{t+5} \right)^3 \), plain differentiation was inadequate. We deftly applied the Chain and Quotient Rules to expertly solve the derivative.

• Chain Rule: Tackles nested functions through a two-step approach by first differentiating the outer function.
• Quotient Rule: Focuses on divisions of two functions, reinforcing precision and organization in calculations.

Mastering these differentiation techniques opens the door to higher-level calculus and problem-solving with ease.