Problem 17
Question
In Problems 1-28, differentiate the functions with respect to the independent variable. $$ g(t)=\left(\frac{t}{t-3}\right)^{3} $$
Step-by-Step Solution
Verified Answer
The derivative of the function is \( g'(t) = \frac{-9t^2}{(t-3)^4} \).
1Step 1: Identify the Function to Differentiate
The function given is \( g(t) = \left( \frac{t}{t-3} \right)^3 \). Here, \( t \) is the independent variable. We need to apply differentiation rules to find the derivative of this function with respect to \( t \).
2Step 2: Use the Chain Rule
We will use the chain rule for differentiation, which is suitable for composite functions. The outer function is \( u^3 \), where \( u = \frac{t}{t-3} \). The derivative of \( u^3 \) with respect to \( u \) is \( 3u^2 \).
3Step 3: Differentiate the Inner Function
The inner function is \( u = \frac{t}{t-3} \). Differentiate this function using the quotient rule. If \( u = \frac{N}{D} \), the quotient rule is \( \frac{d}{dt}\left(\frac{N}{D}\right) = \frac{D \cdot N' - N \cdot D'}{D^2} \). Here, \( N = t \), \( D = t-3 \), \( N' = 1 \), and \( D' = 1 \). Thus, \( \frac{d}{dt}\left(\frac{t}{t-3}\right) = \frac{(t-3)\cdot1 - t\cdot1}{(t-3)^2} = \frac{t-3-t}{(t-3)^2} = \frac{-3}{(t-3)^2} \).
4Step 4: Multiply Using the Chain Rule
According to the chain rule, the derivative \( g'(t) \) is found by multiplying the derivative of the outer function by the derivative of the inner function. Thus, \( g'(t) = 3\left(\frac{t}{t-3}\right)^2 \cdot \frac{-3}{(t-3)^2} \).
5Step 5: Simplify the Derivative
Combine the terms in \( g'(t) \): \( g'(t) = 3 \cdot \frac{-3t^2}{(t-3)^4} \). This simplifies to \( g'(t) = \frac{-9t^2}{(t-3)^4} \).
Key Concepts
Understanding the Chain RuleUsing the Quotient RuleExploring Derivatives
Understanding the Chain Rule
The chain rule is a fundamental principle in calculus that's used when you have to differentiate composite functions.
A composite function is essentially a function within another function, just like nesting dolls. This rule tells us how to find the derivative when functions are inside each other, which is pretty common in calculus problems.
Here's the simple idea: take the derivative of the outer function, then multiply it by the derivative of the inner function. So in our given function, where we have math \( g(t) = \left( \frac{t}{t-3} \right)^3 \), we have an outer function \( (u)^3 \) and an inner function \( u = \frac{t}{t-3} \).
A composite function is essentially a function within another function, just like nesting dolls. This rule tells us how to find the derivative when functions are inside each other, which is pretty common in calculus problems.
Here's the simple idea: take the derivative of the outer function, then multiply it by the derivative of the inner function. So in our given function, where we have math \( g(t) = \left( \frac{t}{t-3} \right)^3 \), we have an outer function \( (u)^3 \) and an inner function \( u = \frac{t}{t-3} \).
- Differentiate the outer function \((u)^3\) with respect to \(u\), yielding \(3u^2\).
- Don't forget to multiply this by the derivative of the inner function \(u\).
Using the Quotient Rule
The quotient rule is your go-to tool when you have to differentiate the division of two functions. This is especially helpful when dealing with ratios or fractions that can't be simplified easily.
When you have a function of the form \( \frac{N}{D} \), where \(N\) is the numerator and \(D\) is the denominator, the quotient rule tells us:
\[\frac{d}{dt}\left(\frac{N}{D}\right) = \frac{D \cdot N' - N \cdot D'}{D^2}\]
In the problem, for \( u = \frac{t}{t-3} \), this rule helps us find the derivative of that inner function. Here's how it works in steps:
When you have a function of the form \( \frac{N}{D} \), where \(N\) is the numerator and \(D\) is the denominator, the quotient rule tells us:
\[\frac{d}{dt}\left(\frac{N}{D}\right) = \frac{D \cdot N' - N \cdot D'}{D^2}\]
In the problem, for \( u = \frac{t}{t-3} \), this rule helps us find the derivative of that inner function. Here's how it works in steps:
- Identify \( N = t \) and \( D = t-3 \).
- Find the derivatives \( N' = 1 \) and \( D' = 1 \).
- Plug them into the quotient rule formula.
Exploring Derivatives
Derivatives are all about understanding how a function changes. Imagine watching a car's speedometer while it accelerates; that speedometer reading is similar to what derivatives can show us about a function.
Derivatives tell us the rate of change of a function concerning its independent variable. In this exercise, we are trying to find how the function \( g(t) = \left( \frac{t}{t-3} \right)^3 \) changes as \( t \) changes.
After tackling both the chain rule and quotient rule steps, we can put everything together to find our answer:
Derivatives tell us the rate of change of a function concerning its independent variable. In this exercise, we are trying to find how the function \( g(t) = \left( \frac{t}{t-3} \right)^3 \) changes as \( t \) changes.
After tackling both the chain rule and quotient rule steps, we can put everything together to find our answer:
- First, find the derivative of the outer function using the chain rule: \(3\left(\frac{t}{t-3}\right)^2\).
- Multiply this by the derivative of the inner function found using the quotient rule: \(\frac{-3}{(t-3)^2}\).
- Simplify to get the final derivative: \( \frac{-9t^2}{(t-3)^4} \).
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