/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 55 \(\begin{array}{l}{\text { (a) U... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\(\begin{array}{l}{\text { (a) Use the Product Rule twice to prove that if } f, g, \text { and } h \text { are }} \\ {\text { differentiable, then }(\mathrm{fgh})^{\prime}=\mathrm{f}^{\prime} g h+\mathrm{f} g^{\prime} h+\mathrm{f} g \mathrm{h}^{\prime} .}\end{array}\) \(\begin{array}{r}{\text { (b) Taking } \mathrm{f}=g=\mathrm{h} \text { in part (a), show that }} \\\ {\frac{\mathrm{d}}{\mathrm{dx}}[\mathrm{f}(\mathrm{x})]^{3}=3[\mathrm{f}(\mathrm{x})]^{2} \mathrm{f}^{\prime}(\mathrm{x})} \\ {\text { (c) Use part (b) to differentiate } \mathrm{y}=\mathrm{e}^{3 \mathrm{x}} \text { . }}\end{array}\)

Short Answer

Expert verified
For part (a), \((fgh)' = f'gh + fg'h + fgh'\); (b) \(\frac{d}{dx}[f(x)]^3 = 3[f(x)]^2 f'(x)\); (c) \(\frac{d}{dx}[e^{3x}] = 3e^{3x}\).

Step by step solution

01

Apply the Product Rule First

The product rule states that if \( u \) and \( v \) are differentiable, then the derivative of their product is \((uv)^{\prime} = u^{\prime}v + uv^{\prime}\). Begin by setting \( u = f \) and \( v = gh \). Then, \((f(gh))^{\prime} = f^{\prime}(gh) + f(gh)^{\prime}\).
02

Apply the Product Rule Second Time

For the term \((gh)^{\prime}\), again apply the product rule with \( u = g \) and \( v = h \). Thus, \((gh)^{\prime} = g^{\prime}h + gh^{\prime}\). Substitute this back into the expression for the derivative from Step 1: \((f(gh))^{\prime} = f^{\prime}gh + f(g^{\prime}h + gh^{\prime})\).
03

Simplify the Expression

Combine the terms by distributing \( f \): \((f(gh))^{\prime} = f^{\prime}gh + fg^{\prime}h + fgh^{\prime}\). This completes the proof that \((fgh)^{\prime} = f^{\prime}gh + fg^{\prime}h + fgh^{\prime}\).
04

Simplify by Setting f = g = h

Let \( f = g = h \). This substitution implies that \( (fgh) = (f^3) \). Therefore, \( (fgh)^{\prime} \) becomes \( (f^3)^{\prime} \). Using the result from Step 3, \((fgh)^{\prime} = f^{\prime}f^2 + ff^{\prime}f + f^2f^{\prime} = 3f^{\prime}f^2\).
05

Differentiate y = e^{3x} Using Part (b)

According to part (b), if \( f(x) = e^x \), the derivative \( (f(x))^3 \) is \( 3(f(x))^2 f^{\prime}(x) \). Therefore, \( \frac{d}{dx}[e^{3x}] = 3(e^x)^2(1)(3e^x) = 3e^{3x} \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Differentiable Functions
Differentiable functions form the heart of calculus, enabling us to find rates of change and solve real-world problems. But what exactly does it mean for a function to be differentiable? Well, if a function is smooth enough without any breaks, jumps, or sharp corners, it is considered differentiable.
Differentiability is defined as having a derivative at every point in its domain. This means the function must have a continuous rate of change. More formally, a function \( f(x) \) is differentiable at a point \( a \) if the following limit exists:
  • \( \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} \).
If this limit exists, then \( f'(a) \) is the derivative of \( f \) at point \( a \). Differentiable functions help us apply calculus tools like the Product Rule, Chain Rule, and more, to analyze complex problems.
Derivative
The concept of a derivative is foundational in calculus. It represents the rate at which a function's value changes as its input changes. In simple terms, the derivative of a function gives us the slope of the tangent line to the function's graph at any point.
Mathematically, the derivative of a function \( f(x) \) at a point \( x \) is denoted as \( f'(x) \), and it can be calculated using the limit:
  • \( f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \).
There are specific rules to compute derivatives, such as the Product Rule, which we saw in the exercise. This rule is crucial when dealing with the product of two or more differentiable functions. If \( u(x) \) and \( v(x) \) are differentiable functions, then their product's derivative is \( (uv)' = u'v + uv' \). Understanding and mastering the derivative concept is key to solving many practical problems involving rates of change.
Chain Rule
The Chain Rule is a powerful tool for differentiating composite functions. Whenever we have a function nested inside another function, the Chain Rule comes into play. In essence, it helps us find the derivative of a composite function by taking the derivative of the outer function and multiplying it by the derivative of the inner function.
The Chain Rule can be written as:
  • Given \( y = f(g(x)) \, \text{the derivative is} \quad \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \).
This rule allows us to efficiently handle derivations that involve complex, multi-layered functions. For example, differentiating \( y = e^{3x} \) requires applying the Chain Rule. The outer function is \( e^{u} \), where \( u = 3x \). We derive the outer function, \( e^{u} \, \) and then multiply it by the derivative of the inner function, \( 3x \. \) Therefore, the derivative is \( 3e^{3x} \. \) Mastery of the Chain Rule is essential for tackling many differentiation problems encountered in calculus.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

$$ \begin{array}{l}{\text { Suppose that we don't have a formula for } g(\mathrm{x}) \text { but we know }} \\ {\text { that } g(2)=-4 \text { and } g^{\prime}(\mathrm{x})=\sqrt{\mathrm{x}^{2}+5} \text { for all } \mathrm{x} \text { . }}\end{array} $$ $$ \begin{array}{l}{\text { (a) Use a linear approximation to estimate } g(1.95)} \\ {\text { and } g(2.05) .} \\ {\text { (b) Are your estimates in part (a) too large or too small? }} \\ {\text { Explain. }}\end{array} $$

The altitude of a triangle is increasing at a rate of 1 \(\mathrm{cm} / \mathrm{min}\) while the area of the triangle is increasing at a rate of 2 \(\mathrm{cm}^{2} / \mathrm{min}\) . At what rate is the base of the triangle changing when the altitude is 10 \(\mathrm{cm}\) and the area is 100 \(\mathrm{cm}^{2} ?\)

Establish the following rules for working with differentials (where c denotes a constant and u and \(v\) are functions of \(x )\) . (a) \(\mathrm{dc}=0\) $$ \mathrm{d}(\mathrm{u}+v)=\mathrm{du}+\mathrm{d} v \quad \text { (d) } \mathrm{d}(\mathrm{u} v)=\mathrm{u} \mathrm{d} v+v $$ $$ \mathrm{d}\left(\frac{\mathrm{u}}{v}\right)=\frac{v \mathrm{du}-\mathrm{u} \mathrm{d} v}{v^{2}} \quad \text { (f) } \mathrm{d}\left(\mathrm{x}^{\mathrm{n}}\right)=\mathrm{nx}^{\mathrm{n}-1} \mathrm{d} \mathrm{x} $$

If \(y = f ( u )\) and \(u = g ( x ) ,\) where \(f\) and \(g\) are twice differen- tiable functions, show that $$\frac { d ^ { 2 } y } { d x ^ { 2 } } = \frac { d ^ { 2 } y } { d u ^ { 2 } } \left( \frac { d u } { d x } \right) ^ { 2 } + \frac { d y } { d u } \frac { d ^ { 2 } u } { d x ^ { 2 } }$$

Two people start from the same point. One walks east at 3 \(\mathrm{mi} / \mathrm{h}\) and the other walks northeast at 2 \(\mathrm{mi} / \mathrm{h} .\) How fast is the distance between the people changing after 15 minutes?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.