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Chapter 3: Problem 73

Suppose that functions \(f\) and \(g\) and their derivatives with respect to \(x\) have the following values at \(x=2\) and \(x=3\) $$\begin{array}{ccccc}\hline x & f(x) & g(x) & f^{\prime}(x) & g^{\prime}(x) \\\\\hline 2 & 8 & 2 & 1 / 3 & -3 \\ 3 & 3 & -4 & 2 \pi & 5 \\\\\hline\end{array}$$ Find the derivatives with respect to \(x\) of the following combinations at the given value of \(x\) a. \(2 f(x), \quad x=2\) b. \(f(x)+g(x), \quad x=3\) c. \(f(x) \cdot g(x), \quad x=3\) d. \(f(x) / g(x), \quad x=2\) e. \(f(g(x)), \quad x=2\) f. \(\sqrt{f(x)}, \quad x=2\) g. \(1 / g^{2}(x), \quad x=3\) h. \(\sqrt{f^{2}(x)+g^{2}(x)}, \quad x=2\)

Short Answer

Expert verified
a. \(\frac{2}{3}\); b. \(2\pi + 5\); c. \(-8\pi + 15\); d. \(6.67\); e. \(-1\); f. \(\frac{1}{12\sqrt{2}}\); g. \(-\frac{5}{32}\); h. \(\frac{-12}{\sqrt{68}}\).

Step by step solution

01

Differentiate 2f(x) at x = 2

The derivative of a constant times a function, like \(2f(x)\), is the constant times the derivative of the function. Therefore, the derivative at \(x = 2\) is: \[ \frac{d}{dx} (2f(x)) = 2 \cdot f'(x) = 2 \cdot \frac{1}{3} = \frac{2}{3}. \]
02

Differentiate f(x) + g(x) at x = 3

The derivative of the sum \(f(x) + g(x)\) is the sum of the derivatives: \[ \frac{d}{dx}(f(x) + g(x)) = f'(x) + g'(x) = 2\pi + 5. \] Calculating this gives: \[ 2\pi + 5. \]
03

Differentiate f(x) 路 g(x) at x = 3 using the Product Rule

The derivative of a product \(f(x) \cdot g(x)\) is given by \(f'(x)g(x) + f(x)g'(x)\). At \(x = 3\): \[ f'(3)g(3) + f(3)g'(3) = 2\pi \cdot (-4) + 3 \cdot 5. \] Simplifying gives \(-8\pi + 15.\)
04

Differentiate \(\frac{f(x)}{g(x)}\) at x = 2 using the Quotient Rule

The derivative of \(\frac{f(x)}{g(x)}\) is \(\frac{f'(x)g(x) - f(x)g'(x)}{(g(x))^2}\). At \(x = 2\): \[ \frac{\frac{1}{3} \cdot 2 - 8 \cdot (-3)}{2^2} = \frac{\frac{2}{3} + 24}{4}. \] Simplifying gives \(\frac{26.67}{4} = 6.67.\)
05

Differentiate f(g(x)) at x = 2 using the Chain Rule

Using the chain rule, the derivative is \(f'(g(x))g'(x)\). At \(x = 2\), evaluate as follows: \[ f'(g(2)) \cdot g'(2) = f'(2) \cdot (-3) = \frac{1}{3} \cdot (-3). \] This results in \(-1.\)
06

Differentiate \(\sqrt{f(x)}\) at x = 2

Using the chain rule for \(\sqrt{f(x)}\), the derivative is \(\frac{1}{2\sqrt{f(x)}} f'(x)\). At \(x = 2\): \[ \frac{1}{2\sqrt{8}} \cdot \frac{1}{3}. \] Simplifying gives \(\frac{1}{12\sqrt{2}}.\)
07

Differentiate \(\frac{1}{g^2(x)}\) at x = 3 using the Chain Rule

This is \((g(x)^{-2})\). The derivative is \(-2g'(x)g(x)^{-3}\), apply at \(x = 3\): \[ -2 \cdot 5 \cdot (-4)^{-3}. \] Calculating gives \(-\frac{5}{32}.\)
08

Differentiate \(\sqrt{f^2(x) + g^2(x)}\) at x = 2 using Chain and Sum Rule

This requires the chain rule: \(\frac{1}{2\sqrt{f^2(x) + g^2(x)}} (2f(x)f'(x) + 2g(x)g'(x))\). At \(x = 2\): \[ \frac{1}{2\sqrt{68}}(2 \cdot 8 \cdot \frac{1}{3} + 2 \cdot 2 \cdot (-3)). \] Simplifying yields \(\frac{-12}{\sqrt{68}}.\)

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Key Concepts

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

Product Rule
The Product Rule is a fundamental technique for finding the derivative of a product of two differentiable functions. When you have a function in the form \[ h(x) = f(x) imes g(x) \] and need to differentiate it, you apply the Product Rule. The derivative, denoted \[ h'(x) \], is given by:
  • Taking the derivative of the first function and multiplying it by the second function.
  • Then, adding the first function times the derivative of the second function.
So, the formula is:\[ h'(x) = f'(x)g(x) + f(x)g'(x) \]This seemingly simple theorem is powerful because real-world functions often come in such multiplicative forms. Ensure to track the derivatives with care to avoid common mistakes!
Chain Rule
The Chain Rule is a core concept in calculus for differentiating compositions of functions. When you have a function like \[ y = f(g(x)) \],this rule effectively helps. It states that the derivative of a composite function is:
  • The derivative of the outer function evaluated at the inner function, times the derivative of the inner function.
Mathematically, the Chain Rule is expressed as: \[ \frac{dy}{dx} = f'(g(x)) imes g'(x) \]This rule is crucial when dealing with more complicated functions that involve nested forms such as trigonometric compositions or square roots of other functions.
Quotient Rule
The Quotient Rule is another key method for differentiation, applied when dealing with a ratio of two differentiable functions. The function form is:\[ h(x) = \frac{f(x)}{g(x)} \]To find the derivative, the Quotient Rule suggests:
  • Multiply the derivative of the numerator by the denominator.
  • Then, subtract the product of the numerator and the derivative of the denominator.
  • Finally, divide the entire expression by the square of the denominator.
The formula looks like this:\[ h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2} \]This rule is indispensable when handling functions that are expressed as one function divided by another. It emphasizes structure over simplicity.
Sum Rule
The Sum Rule in calculus simplifies the process of differentiation when dealing with functions that are added together. If you have functions written as:\[ h(x) = f(x) + g(x) \],then the derivative of their sum is simply:
  • The sum of their derivatives.
Hence, the formula becomes:\[ h'(x) = f'(x) + g'(x) \]This makes computing derivatives easier because you can treat each function individually, compute its derivative, and then combine them. The Sum Rule underscores the idea that differentiation is a linear operation.

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Most popular questions from this chapter

On a morning of a day when the sun will pass directly overhead, the shadow of an \(24 \mathrm{m}\) building on level ground is \(18 \mathrm{m}\) long. At the moment in question, the angle \(\theta\) the sun makes with the ground is increasing at the rate of \(0.27^{\circ} / \mathrm{min} . \mathrm{At}\) what rate is the shadow decreasing? (Remember to use radians. Express your answer in centimeters per minute, to the nearest tenth.)

Use a CAS to perform the following steps a. Plot the equation with the implicit plotter of a CAS. Check to see that the given point \(P\) satisfies the equation. b. Using implicit differentiation, find a formula for the derivative \(d y / d x\) and evaluate it at the given point \(P\). c. Use the slope found in part (b) to find an equation for the tangent line to the curve at \(P .\) Then plot the implicit curve and tangent line together on a single graph. $$x^{3}-x y+y^{3}=7, \quad P(2,1)$$

What happens to the derivatives of \(\sin x\) and \(\cos x\) if \(x\) is measured in degrees instead of radians? To find out, take the following steps. a. With your graphing calculator or computer grapher in degree mode, graph $$f(h)=\frac{\sin h}{h}$$ and estimate \(\lim _{h \rightarrow 0} f(h) .\) Compare your estimate with \(\pi / 180 .\) Is there any reason to believe the limit should be \(\pi / 180 ?\) b. With your grapher still in degree mode, estimate $$\lim _{h \rightarrow 0} \frac{\cos h-1}{h}$$ c. Now go back to the derivation of the formula for the derivative of \(\sin x\) in the text and carry out the steps of the derivation using degree-mode limits. What formula do you obtain for the derivative? d. Work through the derivation of the formula for the derivative of \(\cos x\) using degree-mode limits. What formula do you obtain for the derivative? e. The disadvantages of the degree-mode formulas become apparent as you start taking derivatives of higher order. Try it. What are the second and third degree-mode derivatives of \(\sin x\) and \(\cos x ?\)

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 error \(E(x)=f(x)-g(x)\) 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 \(x-a\)

\(y=\cos x\) for \(-\pi \leq x \leq 2 \pi .\) On the same screen, graph $$y=\frac{\sin (x+h)-\sin x}{h}$$ for \(h=1,0.5,0.3,\) and \(0.1 .\) Then, in a new window, try \(h=-1,-0.5,\) and \(-0.3 .\) What happens as \(h \rightarrow 0^{+2} \mathrm{As} h \rightarrow 0^{-2}\) What phenomenon is being illustrated here?
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