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Chapter 3: Problem 10

In Exercises \(9-18,\) write the function in the form \(y=f(u)\) and \(u=g(x) .\) Then find \(d y / d x\) as a function of \(x .\) $$ y=(4-3 x)^{9} $$

Short Answer

Expert verified
\(\frac{dy}{dx} = -27(4-3x)^8\).

Step by step solution

01

Identify Inner and Outer Functions

We start by identifying the inner and outer functions. In the expression \(y = (4-3x)^9\), the inner function \(u\) is inside the exponentiation and is \(u = 4-3x\). The outer function \(y\) is the exponentiation itself, so \(y = f(u) = u^9\).
02

Rewrite the Function

Rewrite the expression using the identified inner and outer functions. We have \(u = 4 - 3x\) and \(y = f(u) = u^9\). Thus, the original function can be expressed as a composition: \(y = (4-3x)^9 = (u)^9\).
03

Differentiate Outer Function with Respect to Inner Function

Differentiate the outer function \(y = u^9\) with respect to \(u\). Using the power rule, we obtain \(\frac{dy}{du} = 9u^8\).
04

Differentiate Inner Function with Respect to x

Differentiate the inner function \(u = 4 - 3x\) with respect to \(x\). We obtain \(\frac{du}{dx} = -3\).
05

Apply Chain Rule for Derivatives

According to the chain rule, the derivative of \(y\) with respect to \(x\) can be found by multiplying \(\frac{dy}{du}\) by \(\frac{du}{dx}\). So, \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = 9u^8 \cdot (-3)\).
06

Substitute Back the Inner Function

Substitute \(u = 4 - 3x\) back into the expression. Therefore, \(\frac{dy}{dx} = 9(4-3x)^8 \cdot (-3) = -27(4-3x)^8\).

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Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus. It refers to finding the derivative, which represents the rate of change of a function with respect to a variable. In simple terms, differentiation helps us understand how a function changes as its input changes. This is highly useful in various fields such as physics, economics, and biology.

There are several rules and techniques to perform differentiation smoothly. The most basic ones include the power rule, product rule, quotient rule, and chain rule. In our example, we specifically use the power rule and chain rule to find the derivative. Differentiation allows us to find slopes of tangents to curves, velocities from position functions, and many other applications.

Once the derivative is found, it can be used to solve problems like finding maximum and minimum values of functions, optimizing real-world processes, and analyzing the behavior of physical systems. Differentiating a function requires practice and understanding of its rules, but once mastered, it provides powerful tools for analyzing functions.
Composite Functions
Composite functions involve combining two or more functions to produce a new function. In mathematics, a composite function is expressed as \( (f \circ g)(x) = f(g(x)) \). It means applying one function to the results of another function. This concept is essential because it allows us to analyze more complicated functions that cannot be easily handled as single, standalone functions.

In the given exercise, we deal with a composite function where our inner function is \( g(x) = 4 - 3x \), and our outer function is \( f(u) = u^9 \). The original function \( y = (4 - 3x)^9 \) is therefore a composition with \( u \) being \( 4 - 3x \) fed into \( u^9 \).

Understanding composite functions is critical for calculus, as they frequently require the use of the chain rule for differentiation. Composite functions permeate throughout mathematics, and mastery of them bolsters your ability to solve more intricate problems.
Power Rule
The power rule is one of the simplest and most commonly used rules in differentiation. It applies to functions of the form \( x^n \), where \( n \) is any real number. The power rule states that the derivative of \( x^n \) is \( nx^{n-1} \).

In the context of our exercise, the outer function \( y = u^9 \) requires the use of the power rule. We differentiate it with respect to \( u \) to get the derivative \( \frac{dy}{du} = 9u^8 \), as shown when applying the power rule: bring down the exponent and decrease it by one.

The power rule simplifies the process of differentiation significantly, allowing quick computation for polynomial functions. It is widely used because many practical problems involve polynomial expressions. By understanding and practicing the power rule, you gain a powerful tool for handling polynomial differentiation efficiently.

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Most popular questions from this chapter

In Exercises \(67-70\) , use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval \(I\) . Perform the following steps: a. Plot the function \(f\) over \(I .\) b. Find the linearization \(L\) of the function at the point \(a\) . c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta > 0\) as you can, satisfying $$ |x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon $$ for \(\epsilon=0.5,0.1,\) and 0.01 . Then check graphically to see if your \(\delta\) -estimate holds true. $$ f(x)=x^{2 / 3}(x-2), \quad[-2,3], \quad a=2 $$

In Exercises 73 and \(74,\) find both \(d y / d x\) (treating \(y\) as a differentiable function of \(x )\) and \(d x / d y\) (treating \(x\) as a differentiable function of \(y )\) . How do \(d y / d x\) and \(d x / d y\) seem to be related? Explain the relationship geometrically in terms of the graphs. $$ x y^{3}+x^{2} y=6 $$

Normals to a parabola Show that if it is possible to draw three normals from the point \((a, 0)\) to the parabola \(x=y^{2}\) shown here, then \(a\) must be greater than 1\(/ 2\) . One of the normals is the \(x\) -axis. For what value of \(a\) are the other two normals perpendicular?

The linearization is the best linear approximation This is why we use the linearization.) Suppose that \(y=f(x)\) is differentiable at \(x=a\) and that \(g(x)=m(x-a)+c\) is a linear function in which \(m\) and \(c\) are constants. If the erroo \(E(x)=f(x)-g(x)\) were small enough near \(x=a,\) we might think of using \(g\) as a linear approximation of \(f\) instead of the linearization \(L(x)=\) \(f(a)+f^{\prime}(a)(x-a) .\) Show that if we impose on \(g\) the conditions 1\. \(E(a)=0\) 2\. \(\lim _{x \rightarrow a} \frac{E(x)}{x-a}=0\) then \(g(x)=f(a)+f^{\prime}(a)(x-a) .\) Thus, the linearization \(L(x)\) gives the only linear approximation whose error is both zero at \(x=a\) and negligible in comparison with \(x-a\)

In Exercises \(67-70\) , use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval \(I\) . Perform the following steps: a. Plot the function \(f\) over \(I .\) b. Find the linearization \(L\) of the function at the point \(a\) . c. Plot \(f\) and \(L\) together on a single graph. d. Plot the absolute error \(|f(x)-L(x)|\) over \(I\) and find its maximum value. e. From your graph in part (d), estimate as large a \(\delta > 0\) as you can, satisfying $$ |x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon $$ for \(\epsilon=0.5,0.1,\) and 0.01 . Then check graphically to see if your \(\delta\) -estimate holds true. $$ f(x)=x^{3}+x^{2}-2 x, \quad[-1,2], \quad a=1 $$
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