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Chapter 3: Problem 32

Find the derivative of the following functions by first simplifying the expression. \(h(x)=\frac{x^{3}-6 x^{2}+8 x}{x^{2}-2 x}\)

Short Answer

Expert verified
Answer: The derivative of the given function is \(h'(x) = 1\).

Step by step solution

01

Simplify the given function

To simplify the function, we first factor out the common factor from both the numerator and the denominator. $$ h(x) = \frac{x^3 - 6x^2 + 8x}{x^2 - 2x} = \frac{x(x^2 - 6x + 8)}{x(x - 2)} $$ Now, we can further factor the expression in the numerator: $$ h(x) = \frac{x(x - 2)(x - 4)}{x(x - 2)} $$ Cancel out the \((x - 2)\) common factor from the numerator and the denominator: $$ h(x) = \frac{x(x - 4)}{x} $$
02

Apply the quotient rule

Now that the function is simplified, we apply the quotient rule for differentiation to find the derivative of \(h(x)\). The quotient rule states that the derivative of \(\frac{f(x)}{g(x)}\) is: $$ \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} $$ In this case, we have \(f(x) = x(x-4)\) and \(g(x) = x\). To find the respective derivatives, we apply the product rule for the numerator (\(f(x) = x(x-4)\)) and the power rule for the denominator (\(g(x) = x\)). Product rule states that the derivative of \(u(x)v(x)\) is: $$ u'(x)v(x) + u(x)v'(x) $$ So, for \(f(x)\): $$ u(x) = x, \ v(x) = (x - 4), \ u'(x) = 1, \ v'(x) = 1 $$ Applying the product rule: $$ f'(x) = 1(x-4)+x(1) = x - 4 + x = 2x - 4 $$ For \(g(x)\), using the power rule: $$ g'(x) = 1 \times x^{(1-1)} = 1 $$ Now we can apply the quotient rule to find the derivative of \(h(x)\): $$ \begin{aligned} h'(x) & = \frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2} \\ & =\frac{(2x - 4)(x) - x(x-4)(1)}{(x)^2} \\ & = \frac{2x^2 - 4x - x^2 + 4x}{x^2} \\ & = \frac{x^2}{x^2} \end{aligned} $$
03

Simplify the derivative

Finally, we simplify the derivative by canceling out the common factors: $$ h'(x) = \frac{x^2}{x^2} = 1 $$ Thus, the derivative of the given function is \(h'(x) = 1\).

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Key Concepts

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

Quotient Rule
The quotient rule is an essential method in calculus for finding the derivative of a function composed of two differentiable functions in a fraction. It comes into play when you have a function of the form \( \frac{f(x)}{g(x)} \) and you want to differentiate it. The rule itself is expressed as:
  • \( \frac{d}{dx}\left( \frac{f(x)}{g(x)} \right) = \frac{f'(x) g(x) - f(x) g'(x)}{(g(x))^2} \)
This formula shows that to find the derivative, you take:
  • The derivative of the numerator, \( f'(x) \), times the denominator, \( g(x) \)
  • and subtract the numerator, \( f(x) \), times the derivative of the denominator, \( g'(x) \).
  • Then, you divide by the square of the denominator, \( (g(x))^2 \).
Each term has its role and importance. Understanding these helps in correctly applying the quotient rule and avoiding mistakes. Properly applying the quotient rule requires knowing the derivatives of both the numerator and the denominator beforehand.
Product Rule
When dealing with the differentiation of a product of two functions, the product rule is your go-to technique. This rule allows us to find the derivative of the product of two functions, \( u(x) \) and \( v(x) \), and is represented by:
  • \( \frac{d}{dx} \left( u(x) v(x) \right) = u'(x) v(x) + u(x) v'(x) \)
The product rule tells us that the derivative of the first function times the second function, plus the first function times the derivative of the second function, gives the derivative of the entire expression. This approach ensures that each part of the product is taken into account when determining the overall change.Understanding the product rule is vital as it lays the foundation for dealing with more complex scenarios where multiple functions multiply together. It鈥檚 a basic tool in calculus that enhances our ability to analyze function behavior as part of a larger expression, just like in the original problem.
Simplifying Rational Expressions
Simplifying rational expressions is an effective prerequisite step in tackling calculus problems, especially when derivatives are involved. When you have a fraction like \( \frac{x^3 - 6x^2 + 8x}{x^2 - 2x} \), simplifying it can make the differentiation process much more straightforward.To simplify:
  • Factor out common terms in the numerator and the denominator.
  • Cancel any common factors present in both the numerator and denominator.
In the given example, we first factor the expression into \( \frac{x(x - 2)(x - 4)}{x(x - 2)} \). We then cancel the \( (x-2) \) terms, reducing the expression to \( \frac{x(x - 4)}{x} \). This step is crucial as it simplifies the problem, making the application of differentiation rules, like the quotient and product rules, much simpler. Through simplification, the expression becomes more manageable, leading to simpler calculations when finding the derivative.

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Most popular questions from this chapter

Suppose your graphing calculator has two functions, one called sin \(x,\) which calculates the sine of \(x\) when \(x\) is in radians, and the other called \(s(x),\) which calculates the sine of \(x\) when \(x\) is in degrees. a. Explain why \(s(x)=\sin \left(\frac{\pi}{180} x\right)\) b. Evaluate \(\lim _{x \rightarrow 0} \frac{s(x)}{x} .\) Verify your answer by estimating the limit on your calculator.

Proof of \(\frac{d}{d x}(\cos x)=-\sin x\) Use the definition of the derivative and the trigonometric identity $$ \cos (x+h)=\cos x \cos h-\sin x \sin h $$ to prove that \(\frac{d}{d x}(\cos x)=-\sin x\)

The following limits equal the derivative of a function \(f\) at a point a. a. Find one possible \(f\) and \(a\) b. Evaluate the limit. $$\lim _{h \rightarrow 0} \frac{\sin \left(\frac{\pi}{6}+h\right)-\frac{1}{2}}{h}$$

Product Rule for three functions Assume that \(f, g,\) and \(h\) are differentiable at \(x\) a. Use the Product Rule (twice) to find a formula for \(\frac{d}{d x}(f(x) g(x) h(x))\) b. Use the formula in (a) to find \(\frac{d}{d x}\left(e^{2 x}(x-1)(x+3)\right)\)

Orthogonal trajectories Two curves are orthogonal to each other if their tangent lines are perpendicular at each point of intersection (recall that two lines are perpendicular to each other if their slopes are negative reciprocals. . A family of curves forms orthogonal trajectories with another family of curves if each curve in one family is orthogonal to each curve in the other family. For example, the parabolas \(y=c x^{2}\) form orthogonal trajectories with the family of ellipses \(x^{2}+2 y^{2}=k,\) where \(c\) and \(k\) are constants (see figure). Use implicit differentiation if needed to find \(d y / d x\) for each equation of the following pairs. Use the derivatives to explain why the families of curves form orthogonal trajectories. \(x y=a ; x^{2}-y^{2}=b,\) where \(a\) and \(b\) are constants
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