Problem 32
Question
Differentiate $$ f(x)=\frac{3(x-1)^{2}}{2+a} $$ with respect to \(x\). Assume that \(a\) is a positive constant.
Step-by-Step Solution
Verified Answer
The derivative of the function is \( f'(x) = \frac{6(x-1)}{2+a} \).
1Step 1: Identify the Differentiation Rule
In this exercise, we are working with a function in the form of a constant multiplier times a power of a function of x. This suggests the use of the chain rule for differentiation.
2Step 2: Express the Function
The given function is \( f(x) = \frac{3(x-1)^2}{2+a} \). Here, \( \frac{3}{2+a} \) is a constant with respect to \( x \). We can rewrite the function as \( f(x) = C \cdot (x-1)^2 \), where \( C = \frac{3}{2+a} \).
3Step 3: Differentiate Constant Function Multiplication
Apply the chain rule to differentiate \( f(x) \). Since \( C \) is a constant, we have \( \frac{d}{dx}[C \cdot (x-1)^2] = C \cdot \frac{d}{dx}[(x-1)^2] \).
4Step 4: Differentiate the Inner Function
To differentiate \((x-1)^2\), apply the power rule. The derivative of \((x-1)^2\) is \(2(x-1)\cdot \frac{d}{dx}(x-1)\). Since the derivative of \(x-1\) with respect to \(x\) is 1, this becomes \(2(x-1)\cdot 1 = 2(x-1)\).
5Step 5: Multiply by the Constant
Multiply the result from the differentiation of \((x-1)^2\) by the constant \(C\). This gives \( f'(x) = \frac{3}{2+a} \cdot 2(x-1) \).
6Step 6: Simplify the Expression
Combine and simplify the expression to get the final derivative: \( f'(x) = \frac{6(x-1)}{2+a} \).
Key Concepts
Chain RulePower RuleConstant Multiplier Rule
Chain Rule
The Chain Rule is a fundamental tool in calculus for differentiating composite functions. When you have a function inside another function, you need to apply the Chain Rule. It allows you to break down the differentiation process into manageable steps. Imagine you have a ``machine inside a machine'', where the inner machine processes input before the outer machine gives the final output. This is the essence of the Chain Rule.
In mathematical terms, if you have a function \( g(x) \) inside another function \( f(u) \), where \( u = g(x) \), the derivative is given by:
\[ \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \]
In our exercise, the Chain Rule is used because the function \( (x-1)^2 \) is a composition. We take the derivative of the outer function—a power function—and then multiply it by the derivative of the inner function, \( x-1 \). This is how the Chain Rule simplifies complex differentiation.
In mathematical terms, if you have a function \( g(x) \) inside another function \( f(u) \), where \( u = g(x) \), the derivative is given by:
\[ \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \]
In our exercise, the Chain Rule is used because the function \( (x-1)^2 \) is a composition. We take the derivative of the outer function—a power function—and then multiply it by the derivative of the inner function, \( x-1 \). This is how the Chain Rule simplifies complex differentiation.
Power Rule
The Power Rule is one of the simplest and most frequently used differentiation rules. It states that when you have a power function of the form \( x^n \), the derivative is easy to find. This is how it works:
If \( f(x) = x^n \), then \( f'(x) = n \cdot x^{n-1} \).
This formula is intuitive since you're essentially dropping down the exponent as a coefficient and reducing the exponent by one. In our exercise, to differentiate \( (x-1)^2 \), the Power Rule is applied to the outer function, \( x^2 \), yielding:
\[ 2(x-1)^{2-1} = 2(x-1) \].
The Power Rule is used here in conjunction with the Chain Rule. After applying the Power Rule, we multiply by the derivative of the inner function, \( x-1 \), which streamlines our differentiation process.
If \( f(x) = x^n \), then \( f'(x) = n \cdot x^{n-1} \).
This formula is intuitive since you're essentially dropping down the exponent as a coefficient and reducing the exponent by one. In our exercise, to differentiate \( (x-1)^2 \), the Power Rule is applied to the outer function, \( x^2 \), yielding:
\[ 2(x-1)^{2-1} = 2(x-1) \].
The Power Rule is used here in conjunction with the Chain Rule. After applying the Power Rule, we multiply by the derivative of the inner function, \( x-1 \), which streamlines our differentiation process.
Constant Multiplier Rule
The Constant Multiplier Rule states that if you are differentiating a function that's multiplied by a constant, this constant can be factored out of the differentiation process. It significantly simplifies the calculation as you only need to focus on differentiating the function itself.
Mathematically, if you have a function of the form \( C \cdot f(x) \), where \( C \) is a constant, the differentiation is straightforward:
\[ \frac{d}{dx} [C \cdot f(x)] = C \cdot \frac{d}{dx}[f(x)] \].
In our problem, we identify the constant multiplier, \( \frac{3}{2+a} \), and apply the differentiation rules to \( (x-1)^2 \). By using the Constant Multiplier Rule, we factor \( C \) out and just focus on the derivative of the function: \( 2(x-1) \).
The final derivative then is simply \( \frac{3}{2+a} \cdot 2(x-1) \), leading to our simplified answer, \( \frac{6(x-1)}{2+a} \). This rule makes calculations much more efficient, especially in complex functions with constants.
Mathematically, if you have a function of the form \( C \cdot f(x) \), where \( C \) is a constant, the differentiation is straightforward:
\[ \frac{d}{dx} [C \cdot f(x)] = C \cdot \frac{d}{dx}[f(x)] \].
In our problem, we identify the constant multiplier, \( \frac{3}{2+a} \), and apply the differentiation rules to \( (x-1)^2 \). By using the Constant Multiplier Rule, we factor \( C \) out and just focus on the derivative of the function: \( 2(x-1) \).
The final derivative then is simply \( \frac{3}{2+a} \cdot 2(x-1) \), leading to our simplified answer, \( \frac{6(x-1)}{2+a} \). This rule makes calculations much more efficient, especially in complex functions with constants.
Other exercises in this chapter
Problem 32
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