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Chapter 3: Problem 53

In Exercises \(53-58,\) find the value of \((f \circ g)^{\prime}\) at the given value of \(x\) . $$ f(u)=u^{5}+1, \quad u=g(x)=\sqrt{x}, \quad x=1 $$

Short Answer

Expert verified
\((f \circ g)'(1) = \frac{5}{2}\."}

Step by step solution

01

Understand the Function Composition

We are given two functions, \( f(u) = u^5 + 1 \) and \( g(x) = \sqrt{x} \). We are asked to find the derivative of their composition, \((f \circ g)\), at \( x = 1 \). In function composition, \((f \circ g)(x) = f(g(x))\).
02

Express the Composite Function

Substitute \( g(x) = \sqrt{x} \) into \( f(u) \). This gives us:\[ (f \circ g)(x) = f(\sqrt{x}) = (\sqrt{x})^5 + 1 = x^{5/2} + 1 \]
03

Differentiate the Composite Function

Differentiate \( (f \circ g)(x) = x^{5/2} + 1 \) with respect to \( x \). The derivative is:\[ \frac{d}{dx}[x^{5/2} + 1] = \frac{5}{2}x^{3/2} \]
04

Evaluate the Derivative at x=1

Substitute \( x = 1 \) into the derivative:\[ \left. \frac{5}{2}x^{3/2} \right|_{x=1} = \frac{5}{2}(1)^{3/2} = \frac{5}{2}\]
05

Summarize the Result

The derivative of the composite function \( (f \circ g) \) at \( x = 1 \) is \( \frac{5}{2} \).

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Key Concepts

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

Function Composition
Function composition is a fundamental concept in calculus that involves combining two or more functions. When you compose two functions, such as \( f \) and \( g \), you are essentially plugging one function into another. This is represented as \( (f \circ g)(x) = f(g(x)) \).
This means that whatever output we have from \( g(x) \) becomes the input for \( f(u) \). In the given exercise, \( f(u) = u^5 + 1 \) and \( g(x) = \sqrt{x} \), so the composite function is \( (f \circ g)(x) = f(\sqrt{x}) = (\sqrt{x})^5 + 1 \).
  • Think of function composition like using multiple filters on an image; the first filter changes the image and the second develops it further.
  • In mathematical terms, if the function \( g \) modifies \( x \) by some method, then \( f \) further processes \( g(x) \).
Understanding the result of function composition is crucial when progressing to computing derivatives.
Derivative
The derivative of a function is a core concept in calculus that signifies how a function changes when the input changes. In simpler terms, it is the rate of change or the slope of the function at a certain point.
To find the derivative of a composite function like \( (f \circ g)(x) = x^{5/2} + 1 \), you apply the rules of differentiation to each part of the function. Here, we differentiate \( x^{5/2} \) to get \( \frac{5}{2}x^{3/2} \) since the derivative of \( x^n \) is \( nx^{n-1} \).
This portion of a calculus problem helps us understand how outputs vary with slight changes in inputs.
  • Using the power rule is essential here, which states \( \frac{d}{dx} x^n = nx^{n-1} \).
  • Always apply the differentiation rules according to the form of each term in the function.
This step is crucial before moving on to evaluate the derivative at a specific point.
Evaluation at a Point
Once you have the derivative of a function, it's often useful to evaluate that derivative at a specific point. This means you're determining the rate of change of the function at that particular value of \( x \).
In our case, you substitute \( x = 1 \) into the derivative function \( \frac{5}{2}x^{3/2} \) to get \( \frac{5}{2}(1)^{3/2} = \frac{5}{2} \).
Evaluating the derivative provides important information about the behavior of a function, such as its slope, at that specific location.
  • Think of it as zooming in on the function graph at \( x = 1 \) to see how steep it is there.
  • It also tells us whether the function is increasing or decreasing at \( x = 1 \).
By interpreting the derivative value, we can make conclusions about the function's dynamics at specific points.

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Most popular questions from this chapter

Temperature and the period of a pendulum For oscillations of small amplitude (short swings), we may safely model the relationship between the period \(T\) and the length \(L\) of a simple pendulum with the equation $$ T=2 \pi \sqrt{\frac{L}{g}} $$ where \(g\) is the constant acceleration of gravity at the pendulum's location. If we measure \(g\) in centimeters per second squared, we measure \(L\) in centimeters and \(T\) in seconds. If the pendulum is made of metal, its length will vary with temperature, either increasing or decreasing at a rate that is roughly proportional to \(L .\) In symbols, with \(u\) being temperature and \(k\) the proportionality constant, $$ \frac{d L}{d u}=k L $$ Assuming this to be the case, show that the rate at which the pe- tiod changes with respect to temperature is \(k T / 2\) .

In Exercises \(19-30,\) find \(d y\) $$ 2 y^{3 / 2}+x y-x=0 $$

Which of the following could be true if \(f^{\prime \prime}(x)=x^{-1 / 3} ?\) $$ \begin{array}{ll}{\text { a. } f(x)=\frac{3}{2} x^{2 / 3}-3} & {\text { b. } f(x)=\frac{9}{10} x^{5 / 3}-7} \\ {\text { c. } f^{\prime \prime \prime}(x)=-\frac{1}{3} x^{-4 / 3}} & {\text { d. } f^{\prime}(x)=\frac{3}{2} x^{2 / 3}+6}\end{array} $$

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. $$ x y^{3}+x^{2} y=6 $$

In Exercises 45 and \(46,\) find the slope of the curve at the given points. $$ \left(x^{2}+y^{2}\right)^{2}=(x-y)^{2} \quad \text { at } \quad(1,0) \text { and }(1,-1) $$
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