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Chapter 8: Problem 16

Differentiate (a) \(1 /(x+3)^{2}\) (b) \(\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^{2}\) (c) \(x / \sqrt{\left(x^{2}-1\right)}\) (d) \((2 x+1)^{2} /\left(3 x^{2}+1\right)^{3}\)

Short Answer

Expert verified
(a) \( -\frac{2}{(x+3)^3} \); (b) \( \frac{1}{\sqrt{x}} - \frac{1}{x^{3/2}} \); (c) \( \frac{1}{(x^2 - 1)^{3/2}} \); (d) complex expression.

Step by step solution

01

Differentiate Part (a)

To differentiate the function \( f(x) = \frac{1}{{(x+3)}^2} \), we use the power rule combined with the chain rule.1. Rewrite \( f(x) \) as \( (x+3)^{-2} \).2. Use the chain rule: If \( g(x) = (x+3)^{-2} \), then \( g'(x) = -2(x+3)^{-3} \cdot (1) \).So, the derivative is \( f'(x) = -\frac{2}{(x+3)^3} \).
02

Differentiate Part (b)

For the function \( f(x) = \left( \sqrt{x} + \frac{1}{\sqrt{x}} \right)^2 \), use the chain rule and power rule.1. Let \( u = \sqrt{x} + \frac{1}{\sqrt{x}} \), then \( f(x) = u^2 \).2. Differentiate \( f(x) = u^2 \) using the power rule: \( \frac{d}{dx}[u^2] = 2u \cdot \frac{du}{dx} \).3. Compute \( \frac{du}{dx} = \frac{1}{2\sqrt{x}} - \frac{1}{2x^{3/2}} \).4. Substitute back: \( f'(x) = 2\left( \sqrt{x} + \frac{1}{\sqrt{x}} \right) \left( \frac{1}{2\sqrt{x}} - \frac{1}{2x^{3/2}} \right) \).Simplify to get: \( f'(x) = \frac{1}{\sqrt{x}} - \frac{1}{x^{3/2}} \).
03

Differentiate Part (c)

For \( f(x) = \frac{x}{\sqrt{x^2 - 1}} \), apply the quotient and chain rules.1. Let \( u = x \) and \( v = \sqrt{x^2 - 1} \), then \( f(x) = \frac{u}{v} \).2. Differentiate using quotient rule: \( f'(x) = \frac{v \cdot u' - u \cdot v'}{v^2} \), where \( u' = 1 \).3. For \( v \), \( v' = \frac{1}{2\sqrt{x^2 - 1}} \cdot 2x = \frac{x}{\sqrt{x^2 - 1}} \).4. Substitute: \( f'(x) = \frac{\sqrt{x^2 - 1} - x \cdot \frac{x}{\sqrt{x^2 - 1}}}{x^2 - 1} \).Simplify: \( f'(x) = \frac{1}{(x^2 - 1)^{3/2}} \).
04

Differentiate Part (d)

For \( f(x) = \frac{(2x+1)^2}{(3x^2+1)^3} \), use the quotient rule and chain rule.1. Let \( u = (2x+1)^2 \) and \( v = (3x^2+1)^3 \).2. Differentiate \( u \, (u') \) using the chain rule: If \( u = (2x+1)^2 \), then \( u' = 2(2x+1) \cdot 2 = 4(2x+1) \).3. Differentiate \( v \, (v') \) using the chain rule: If \( v = (3x^2+1)^3 \), then \( v'= 3(3x^2+1)^2 \cdot 6x = 18x(3x^2+1)^2 \).4. Use quotient rule: \( f'(x) = \frac{v\cdot u' - u\cdot v'}{v^2} \).5. Substitute and simplify: \ - \( f'(x) = \frac{(3x^2+1)^3 \cdot 4(2x+1) - (2x+1)^2 \cdot 18x(3x^2+1)^2 }{(3x^2+1)^6} \).Further simplification gives \( f'(x) \) in terms of reduced fractions.

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Key Concepts

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

Power Rule
In calculus, the Power Rule is a straightforward method used to differentiate functions of the form \( x^n \). This means, if you have a function \( f(x) = x^n \), its derivative is \( f'(x) = nx^{n-1} \).

This rule makes it easy to find derivatives whenever you have pure power functions of \( x \), such as \( x^2 \), \( x^3 \), and even \( x^{-2} \). For example, if you want to differentiate \( f(x) = x^3 \), using the power rule would yield \( f'(x) = 3x^2 \).

It simplifies differentiation by reducing steps, but remember, each time you apply this rule, both the original exponent and the newly adjusted one parameterize how the slope of the curve changes.
Chain Rule
The Chain Rule is a method for finding the derivative of composite functions. A composite function is where one function is inside of another, such as \( f(g(x)) \).

To apply the chain rule, differentiate the outer function and multiply by the derivative of the inner function. For example, with \( f(x) = (x^2 + 1)^3 \), consider \( g(x) = x^2 + 1 \) and \( h(g) = g^3 \). Differentiate \( h(g) = g^3 \) first, resulting in \( 3g^2 \), then differentiate \( g(x) = x^2 + 1 \) to get \( 2x \).

Multiply these derivatives: \( 3(x^2 + 1)^2 \times 2x \). This is the essence of the chain rule: take it step-by-step, one layer at a time.
Quotient Rule
The Quotient Rule is used when you need to find the derivative of a function that is the ratio of two differentiable functions. If \( f(x) = \frac{u(x)}{v(x)} \), the derivative \( f'(x) \) is
  1. Calculate derivative of the numerator, \( u'(x) \).
  2. Calculate derivative of the denominator, \( v'(x) \).
  3. Apply: \( f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2} \).
For example, differentiating \( \frac{x}{x^2+1} \), set \( u = x \), \( v = x^2+1 \). Then, \( u' = 1 \), \( v' = 2x \). Substitute and simplify: \( f'(x) = \frac{(x^2+1)(1) - x(2x)}{(x^2+1)^2} \).

This rule highlights the importance of subtracting the product of cross derivatives, ensuring precision in calculations.
Mathematical Functions
Understanding Mathematical Functions is vital for differentiation. Functions map inputs (often \( x \)) to outputs (\( f(x) \)) and can be polynomial, exponential, trigonometric, logarithmic, among others.
  • Polynomials: E.g., \( f(x) = x^2 + 3x + 2 \).
  • Rational: E.g., \( f(x) = \frac{1}{x} \).
  • Trigonometric: E.g., \( f(x) = \sin(x) \).
Each type poses unique challenges and often requires distinct differentiation techniques.

By mastering these, complex derivations become manageable, allowing you to see how small changes in \( x \) affect \( f(x) \). Real-world modeling, signal processing, and optimization rely heavily on these analyses, illustrating the broad application of mathematical functions beyond theoretical calculus.

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Most popular questions from this chapter

Consider the function \(f(x)=2 x^{3}-3 x^{2}+x+3\). Find (a) the derivative of \(f(x)\) from first principles; (b) the rate of change of \(f(x)\) at \(x=1\); (c) the points at which the line through \((1,3)\) with slope \(m\) cuts the graph of \(f(x)\);(d) the values of \(m\) such that two of the points of intersection found in \((\mathrm{c})\) are coincident; (c) the equations of the tangents to the graph of \(f(x)\) at \(x=1\) and \(x=\frac{1}{4}\)

Draw the graph of the function \(f(x)=2 x-1\) for \(-3

Find the equations of the tangent and normal to the curve having equation $$ y^{2}-2 y-4 x+1=0 $$ at the point \((1,3)\).

A lecture theatre having volume \(1000 \mathrm{~m}^{3}\) is designed to seat 200 people. The air is conditioned continuously by an inflow of fresh air at a constant rate \(V\left(\right.\) in \(\mathrm{m}^{3} \mathrm{~min}^{-1}\) ). An average person generates \(980 \mathrm{~cm}^{3}\) of \(\mathrm{CO}_{2}\) per minute, while fresh air contains \(0.04 \%\) of \(\mathrm{CO}_{2}\) by volume. Show that the percentage concentration \(x\) of \(\mathrm{CO}_{2}\) by volume in the lecture theatre at time \(t\) (in min) after the audience enters satisfies the differential equation $$ 1000 \frac{\mathrm{d} x}{\mathrm{~d} t}=19.6+0.04 V-V x(t) $$

The area enclosed between the \(x\) axis, the curve \(y=x(2-x)\) and the ordinates \(x=1\) and \(x=2\) is rotated through \(2 \pi\) radians about the \(x\) axis. Calculate. (a) the rotating area and the coordinates of its centroid: (b) the volume of the solid of revolution formed and the coordinates of its centre of gravity.
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