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Chapter 4: Problem 45

Differentiate the functions in Problems 1-52 with respect to the independent variable. $$ f(x)=2 \sqrt{x^{2}-1} $$

Short Answer

Expert verified
The derivative of \( f(x) = 2\sqrt{x^2 - 1} \) is \( \frac{2x}{\sqrt{x^2 - 1}} \).

Step by step solution

01

Identify the function structure

The given function is \( f(x) = 2\sqrt{x^2 - 1} \). Notice that it is a composite function with the form \( g(h(x)) \), where \( g(u) = 2\sqrt{u} \) and \( h(x) = x^2 - 1 \). We will need to apply the chain rule to differentiate this function.
02

Differentiate the outer function

Differentiate the outer function \( g(u) = 2\sqrt{u} \) with respect to \( u \). The derivative \( g'(u) = \frac{d}{du}(2u^{1/2}) = 2 \cdot \frac{1}{2} u^{-1/2} = u^{-1/2} = \frac{1}{\sqrt{u}} \).
03

Differentiate the inner function

Find the derivative of the inner function \( h(x) = x^2 - 1 \) with respect to \( x \). The derivative \( h'(x) = \frac{d}{dx}(x^2 - 1) = 2x \).
04

Apply the chain rule

Use the chain rule, which states \( \frac{d}{dx} f(x) = g'(h(x)) \times h'(x) \). Substituting the derivatives calculated, we have \( \frac{d}{dx} f(x) = \frac{1}{\sqrt{x^2 - 1}} \times 2x \).
05

Simplify the derivative

Simplify the expression from applying the chain rule. The derivative \( \frac{d}{dx} f(x) = \frac{2x}{\sqrt{x^2 - 1}} \).

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Key Concepts

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

Chain Rule
The chain rule is an essential tool in calculus, used to differentiate composite functions. It lets us find the derivative of functions formed by other functions. Suppose we have a function made from two functions like \( g(h(x)) \). To differentiate this, imagine peeling back layers.
  • First, compute the derivative of the outer function \( g(u) \), assuming \( u = h(x) \).
  • Next, differentiate the inner function \( h(x) \).
Combine these derivatives according to the chain rule:\[\frac{d}{dx} f(x) = g'(h(x)) \cdot h'(x)\]Think of the chain rule as working through paths, starting from the outer layer and moving inward, capturing changes at each level.Deploying this method is powerful for intricate functions where direct differentiation is cumbersome.
Composite Functions
Composite functions join two or more functions into a single expression. They often appear in the structure \( g(h(x)) \), where one function, \( h(x) \), is nested inside another, \( g(u) \). Understanding composite functions is crucial for applying the chain rule efficiently.
A good way to grasp this concept is by:
  • Identifying the outer function \( g(u) \), where \( u = h(x) \).
  • Then, spotting the inner function \( h(x) \).
Composite functions are common in real-world applications, like calculating the interest compounded over time or the temperature change based on various atmospheric effects.
Recognizing these layers is key to unraveling complex equations and making the differentiation process systematic and straightforward.
Derivative Computation
Computing derivatives involves determining how a function changes as its input changes. For composite functions, translating these changes through multiple layers can seem tricky, but it's manageable with the chain rule.
In derivative computation:
  • Identify layers of functions (outer and inner).
  • Apply rules: power rule for simple polynomials, product, and quotient rules for attacking more elaborate constructs.
  • Simplify at each step for clarity and ease of interpretation.
For the function in our exercise, \( f(x) = 2\sqrt{x^2 - 1} \), we first differentiated \( g(u) = 2\sqrt{u} \) into \( \frac{1}{\sqrt{u}} \), then found the derivative of \( h(x) = x^2 - 1 \) as \( 2x \).
Employing the chain rule, we get \( \frac{2x}{\sqrt{x^2 - 1}} \) as the derivative, illustrating effective derivative computation.

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Most popular questions from this chapter

Approximate \(f(x)\) at a by the linear approximation $$L(x)=f(a)+f^{\prime}(a)(x-a)$$ \(f(x)=(1-x)^{-n}\) at \(a=0\). (Assume that \(n\) is a positive integer.)

Use the formula $$f(x) \approx f(a)+f^{\prime}(a)(x-a)$$ to approximate the value of the given function. Then compare your result with the value you get from a calculator. $$ e^{0.1} $$

Use the formula $$f(x) \approx f(a)+f^{\prime}(a)(x-a)$$ to approximate the value of the given function. Then compare your result with the value you get from a calculator. $$ \sin \left(\frac{\pi}{2}+0.02\right) $$

A measurement error in \(x\) affects the accuracy of the value \(f(x) .\) In each case, determine an interval of the form $$[f(x)-\Delta f, f(x)+\Delta f]$$ that reflects the measurement error \(\Delta x .\) In each problem, the quantities given are \(f(x)\) and \(x=\) true value of \(x \pm|\Delta x|\). $$ f(x)=e^{x}, x=2 \pm 0.2 $$

Assume that a species lives in a habitat that consists of many islands close to a mainland. The species occupies both the mainland and the islands, but, although it is present on the mainland at all times, it frequently goes extinct on the islands. Islands can be recolonized by migrants from the mainland. The following model keeps track of the fraction of islands occupied: Denote the fraction of islands occupied at time \(t\) by \(p(t)\). Assume that each island experiences a constant risk of extinction and that vacant islands (the fraction \(1-p\) ) are colonized from the mainland at a constant rate. Then $$ \frac{d p}{d t}=c(1-p)-e p $$ where \(c\) and \(e\) are positive constants. (a) The gain from colonization is \(f(p)=c(1-p)\) and the loss from extinction is \(g(p)=e p .\) Graph \(f(p)\) and \(g(p)\) for \(0 \leq p \leq 1\) in the same coordinate system. Explain why the two graphs intersect whenever \(e\) and \(c\) are both positive. Compute the point of intersection and interpret its biological meaning. (b) The parameter \(c\) measures how quickly a vacant island becomes colonized from the mainland. The closer the islands, the larger is the value of \(c\). Use your graph in (a) to explain what happens to the point of intersection of the two lines as \(c\) increases. Interpret your result in biological terms.
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