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Chapter 3: Problem 27

Differentiate the function. \(f(x)=k(a-x)(b-x)\)

Short Answer

Expert verified
The derivative is \( f'(x) = k(2x - (a+b)) \).

Step by step solution

01

Expand the Function

The first step is to expand the given function before differentiating. The function is \( f(x) = k(a-x)(b-x) \). Use the distributive property to expand it.\[ f(x) = k((a-x)(b-x)) = k((ab - xb - xa + x^2)) = k(x^2 - (a + b)x + ab) \]
02

Differentiate the Expanded Function

Now, differentiate the expanded function term by term with respect to \( x \). The function is \( f(x) = k(x^2 - (a+b)x + ab) \).To differentiate: 1. Apply the power rule to \( x^2 \).2. Apply the constant multiplication rule.The derivative is:\[ f'(x) = k(2x - (a+b)) \]
03

Simplify the Derivative

In this last step, simplify the expression for the derivative if necessary. The expression we derived is already simplified: \[ f'(x) = k(2x - (a+b)) \]This is the final derivative of the function.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Expanded Function
When we talk about expanding a function, we're essentially looking to transform it into a more manageable form for operations like differentiation. In our example, the function is \( f(x) = k(a-x)(b-x) \).
Expanding means using algebraic manipulation to simplify or reorganize it.

Here's how we do it:
  • Take the expression \( (a-x)(b-x) \) and apply the distributive property.
  • This means multiplying each term within the first set of parentheses by each term in the second set, resulting in \( ab - xb - xa + x^2 \).
This expansion makes it easier to differentiate, as we now have a straightforward polynomial \( k(x^2 - (a+b)x + ab) \).
This step is crucial because differentiating simple terms is much easier than handling complex expressions.
Power Rule
The power rule is a fundamental differentiation technique used to find the derivative of terms of the form \( x^n \). For any real number \( n \), the power rule states:
For \( y = x^n \), the derivative \( y' = nx^{n-1} \).

Applying this concept, in our example:
  • The term \( x^2 \) in the expanded function becomes \( 2x^{2-1} = 2x \) when derived.
The power rule is extremely efficient, making it one of the easiest rules to apply, especially when differentiating polynomials.
Remember, if you have a constant in front (like in our function, where 2 is multiplied by \( x \)), it remains in the final derivative.
Distributive Property
The distributive property is an algebraic rule that allows us to multiply a single term by a set of terms added together and distribute the multiplication over each term individually. In essence, for terms \( a, b, c \), and \( d \):
\( a(b+c) = ab + ac \)

In the context of our function, we applied this to \( (a-x)(b-x) \):
  • Multiplied \( (a-x) \) by each term in \( (b-x) \).
  • This produced the expanded form \( ab - xb - xa + x^2 \).
This step revealed each individual term, making it easier to apply differentiation rules.
The distributive property helps us unravel complex expressions into simpler pieces.
Constant Multiplication Rule
The constant multiplication rule in differentiation is simple yet powerful.
It states that if you have a constant multiplied by a function, the derivative of the whole function is simply the constant multiplied by the derivative of the function. Mathematically:
If \( y = c \, g(x) \), then \( y' = c \, g'(x) \).

In our function \( f(x) = k(x^2 - (a+b)x + ab) \):
  • Each term that we derived from applying the power rule is multiplied by the constant \( k \).
  • The derivative becomes \( f'(x) = k(2x - (a+b)) \).
This rule simplifies the calculation by allowing the constant to "pass through" the differentiation process unchanged.

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Most popular questions from this chapter

Find the derivative of the function using the definition of a derivative. State the domain of the function and the domain of its derivative. \(f(x)=x+\sqrt{x}\)

(a) Use the definition of the derivative to calculate \(f^{\prime}\) , (b) Check to see that your answer is reasonable by comparing the graphs of \(f\) and \(f^{\prime}\) \(f(t)=t^{2}-\sqrt{t}\)

Find \(d y / d x\) by implicit differentiation. \(e^{y} \cos x=1+\sin (x y)\)

Find the derivative of the function using the definition of a derivative. State the domain of the function and the domain of its derivative. \(f(t)=5 t-9 t^{2}\)

(a) Find \(y^{\prime}\) by implicit differentiation. (b) Solve the equation explicitly for \(y\) and differentiate to get \(y^{\prime}\) in terms of \(x .\) (c) Check that your solutions to parts (a) and (b) are consistent by substituting the expression for \(y\) into your solution for part (a). \(\cos x+\sqrt{y}=5\)
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