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Chapter 1: Problem 8

Differentiate two ways: first, by using the Product Rule; then, by multiplying the expressions before differentiating. Compare your results as a check. Use a graphing calculator to check your results. \(y=(4 \sqrt{x}+3) x^{3}\)

Short Answer

Expert verified
Both methods of differentiation yield the same derivative: \( y' = 14x^{2.5} + 9x^2 \).

Step by step solution

01

Understanding the Given Expression

We are given the function \( y = (4 \sqrt{x} + 3)x^3 \). Our task is to find its derivative using two methods: the Product Rule and expansion by multiplying before differentiating.
02

Product Rule Approach

The derivative of a product of two functions \( u \) and \( v \) is given by \( (uv)' = u'v + uv' \). Identify \( u = 4\sqrt{x} + 3 \) and \( v = x^3 \). Compute \( u' \) and \( v' \):\[ u' = \frac{d}{dx}(4\sqrt{x} + 3) = \frac{4}{2\sqrt{x}} = \frac{2}{\sqrt{x}}, \ v' = \frac{d}{dx}(x^3) = 3x^2. \] Apply the product rule: \[ y' = (4\sqrt{x} + 3)' x^3 + (4\sqrt{x} + 3) (x^3)'. \] Substitute the derivatives: \[ y' = \left(\frac{2}{\sqrt{x}}\right)x^3 + (4\sqrt{x} + 3)(3x^2). \] Simplify: \[ y' = 2x^{5/2} + (12x^{7/2} + 9x^2). \]
03

Expand before Differentiating

Expand \( y = (4\sqrt{x} + 3)x^3 \) by multiplying: \[ y = 4\sqrt{x} \cdot x^3 + 3x^3 = 4x^{3.5} + 3x^3 \]. Differentiate this expression: \[ \frac{d}{dx}(y) = \frac{d}{dx}(4x^{3.5} + 3x^3) = 4 \cdot 3.5 \cdot x^{2.5} + 3 \cdot 3 \cdot x^2 = 14x^{2.5} + 9x^2. \]
04

Compare the Results

The derivative using the Product Rule simplifies to \( y' = 2x^{5/2} + 12x^{7/2} + 9x^2 \). The expanded expression's derivative is \( y' = 14x^{2.5} + 9x^2 \). Notice that \( 12x^{5/2} + 2x^{5/2} = 14x^{5/2} \) which matches the expansion approach, confirming the results are consistent.
05

Verify with a Graphing Calculator

Input both derivative expressions into a graphing calculator. Plot them and verify that the graphs are identical, thus confirming that both differentiation methods yield the same result.

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Key Concepts

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

Product Rule
The product rule is a fundamental tool in calculus for finding the derivative of the product of two functions. If you have a function defined as a product, say\( u(x) \) and \( v(x) \), the derivative is given by:
  • \((uv)' = u' \, v + u \, v'\)
This formula says: "differentiate the first function and multiply it by the second function, then differentiate the second function and multiply it by the first, and finally add the two products together."
In our example, \( y = (4\sqrt{x} + 3) x^3 \), we first identify \( u = 4\sqrt{x} + 3 \) and \( v = x^3 \). We then find their derivatives as follows:
  • \( u' = \frac{2}{\sqrt{x}} \)
  • \( v' = 3x^2 \)
Applying the product rule, we get:
  • \( y' = \left(\frac{2}{\sqrt{x}}\right) x^3 + (4\sqrt{x} + 3)(3x^2) \)
After some algebraic simplification, the result is:
  • \( y' = 2x^{5/2} + 12x^{7/2} + 9x^2 \)
Function Expansion
Function expansion is a technique where you simplify a function by expanding the expression before differentiating. This can often make the derivative easier to find and understand.
In the problem \( y = (4\sqrt{x} + 3)x^3 \), we expand the expression by multiplying terms directly:
  • \( y = 4x^{3.5} + 3x^3 \)
This expansion translates the original product into simpler components which can be differentiated separately.
Next, differentiate each term individually:
  • \( \frac{d}{dx}(4x^{3.5}) = 14x^{2.5} \)
  • \( \frac{d}{dx}(3x^3) = 9x^2 \)
Thus, the derivative becomes:
  • \( y' = 14x^{2.5} + 9x^2 \)
This demonstrates the comparative ease of finding derivatives by expanding the function first.
Derivative Verification
After finding derivatives using different methods, it's crucial to make sure that results are consistent. This process of checking derivatives is known as derivative verification.
In this example, we've derived \( y' = 2x^{5/2} + 12x^{7/2} + 9x^2 \) using the product rule and \( y' = 14x^{2.5} + 9x^2 \) from function expansion. To confirm their equivalency, simplify and compare the results:
  • Observe that \( 2x^{5/2} + 12x^{7/2} = 14x^{5/2} \), which matches the result from expansion.
  • Both results include \( 9x^2 \) as well.
These steps verify that both differentiation methods are accurate and yield consistent results.
Graphing Calculator Usage
A graphing calculator is a powerful tool to visually verify calculus problems by graphing the original and derivative functions. When you input these functions into the calculator, you can see if their graphs match exactly.
Here's how to use it for this exercise:
  • Input the original function \( y = (4\sqrt{x} + 3)x^3 \).
  • Input both derived expressions: \( y' = 2x^{5/2} + 12x^{7/2} + 9x^2 \) and \( y' = 14x^{2.5} + 9x^2 \).
  • Check that the graphs of both derivative expressions overlay each other perfectly, confirming they represent the same function graphically.
This graphical verification serves as a visual proof of your algebraic differentiation, providing confidence in your results.

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Most popular questions from this chapter

At the beginning of a trip, the odometer on a car reads \(30,680,\) and the car has a full tank of gas. At the end of the trip, the odometer reads \(31,077 .\) It takes 13.5 gal of gas to refill the tank. a) What is the average rate at which the car was traveling, in miles per gallon? b) What is the average rate of gas consumption in gallons per mile?

For each function, find the points on the graph at which the tangent line has slope 1 . $$ y=-0.025 x^{2}+4 x $$

For each function, find the points on the graph at which the tangent line is horizontal. If none exist, state that fact. $$ y=x^{3}-6 x+1 $$

Let \(f\) and \(g\) be differentiable over an open interval containing \(x=a\). If $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{0}{0} \quad \text { or } \quad \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{\pm \infty}{\pm \infty} $$ and if \(\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right)\) exists, then $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right) $$ The forms \(0 / 0\) and \(\pm \infty / \pm \infty\) are said to be indeterminate. In such cases, the limit may exist, and l'H么pital's Rule offers a way to find the limit using differentiation. For example, in Example 1 of Section \(1.1,\) we showed that $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=2 $$ Since, for \(x=1,\) we have \(\left(x^{2}-1\right)(x-1)=0 / 0,\) we differentiate the numerator and denominator separately, and reevaluate the limit: $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=\lim _{x \rightarrow 1}\left(\frac{2 x}{1}\right)=2 $$ Use this method to find the following limits. Be sure to check that the initial substitution results in an indeterminate form. $$ \lim _{x \rightarrow 1}\left(\frac{x^{3}+2 x-3}{x^{2}-1}\right) $$

Use GRAPH and TRACE to find each limit. When necessary, state that the limit does not exist. $$ \begin{array}{l} \text { For } g(x)=\frac{20 x^{2}}{x^{3}+2 x^{2}+5 x} \\ \text { find } \lim _{x \rightarrow \infty} g(x) \text { and } \lim _{x \rightarrow-\infty} g(x) . \end{array} $$
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