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Quotient Rule - Calculus 1

  1. Let h(x)=1+sin⁡xcos⁡xh(x) = \frac{1+\sin x}{\cos x}. Which of the following is an expression for h′(x)h'(x)?
    Answer: 11−sin⁡x\frac{1}{1-\sin x} We use the Quotient Rule with f(x)=1+sin⁡xf(x)=1+\sin x and g(x)=cos⁡xg(x)=\cos x. We have f′(x)=cos⁡xf'(x)=\cos x and g′(x)=−sin⁡xg'(x)=-\sin x. h′(x)=(cos⁡x)(cos⁡x)−(1+sin⁡x)(−sin⁡x)(cos⁡x)2h'(x) = \frac{(\cos x)(\cos x) - (1+\sin x)(-\sin x)}{(\cos x)^2} h′(x)=cos⁡2x+sin⁡x+sin⁡2xcos⁡2xh'(x) = \frac{\cos^2 x + \sin x + \sin^2 x}{\cos^2 x} Using the identity sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1, the numerator becomes 1+sin⁡x1+\sin x. h′(x)=1+sin⁡xcos⁡2xh'(x) = \frac{1+\sin x}{\cos^2 x} Using the identity cos⁡2x=1−sin⁡2x\cos^2 x = 1 - \sin^2 x, we can rewrite the denominator. h′(x)=1+sin⁡x1−sin⁡2xh'(x) = \frac{1+\sin x}{1-\sin^2 x} Factoring the denominator as a difference of squares, (1−sin⁡x)(1+sin⁡x)(1-\sin x)(1+\sin x): h′(x)=1+sin⁡x(1−sin⁡x)(1+sin⁡x)=11−sin⁡xh'(x) = \frac{1+\sin x}{(1-\sin x)(1+\sin x)} = \frac{1}{1-\sin x}
  2. For what non-zero value of xx does the function f(x)=x2x−1f(x) = \frac{x^2}{x-1} have a horizontal tangent line?
    Answer: x=2x = 2 A horizontal tangent line occurs where the derivative is zero. First, find the derivative using the quotient rule: f′(x)=(2x)(x−1)−(x2)(1)(x−1)2=2x2−2x−x2(x−1)2=x2−2x(x−1)2f'(x) = \frac{(2x)(x-1) - (x^2)(1)}{(x-1)^2} = \frac{2x^2 - 2x - x^2}{(x-1)^2} = \frac{x^2 - 2x}{(x-1)^2}. Set f′(x)=0f'(x) = 0, which means the numerator must be zero: x2−2x=0  ⟹  x(x−2)=0x^2 - 2x = 0 \implies x(x-2) = 0. The solutions are x=0x=0 and x=2x=2. The question asks for the non-zero value, which is x=2x=2.
  3. What is the equation of the normal line to the graph of f(x)=exx2+1f(x) = \frac{e^x}{x^2+1} at the point where x=1x=1?
    Answer: x=1x=1 First, find the point of tangency by evaluating f(1)=e112+1=e2f(1) = \frac{e^1}{1^2+1} = \frac{e}{2}. The point is (1,e2)(1, \frac{e}{2}). Next, find the slope of the tangent line by finding the derivative f′(x)f'(x) using the Quotient Rule. f′(x)=(x2+1)(ex)−(ex)(2x)(x2+1)2=ex(x2−2x+1)(x2+1)2=ex(x−1)2(x2+1)2f'(x) = \frac{(x^2+1)(e^x) - (e^x)(2x)}{(x^2+1)^2} = \frac{e^x(x^2 - 2x + 1)}{(x^2+1)^2} = \frac{e^x(x-1)^2}{(x^2+1)^2} Now, evaluate the slope at x=1x=1: f′(1)=e1(1−1)2(12+1)2=e(0)222=0f'(1) = \frac{e^1(1-1)^2}{(1^2+1)^2} = \frac{e(0)^2}{2^2} = 0 A slope of 0 means the tangent line is horizontal. The equation of the tangent line is y=e2y = \frac{e}{2}. The normal line is perpendicular to the tangent line. A line perpendicular to a horizontal line is a vertical line. The equation of the vertical line passing through the point (1,e2)(1, \frac{e}{2}) is x=1x=1.
  4. Let h(x)=f(x)g(x)h(x) = \frac{f(x)}{g(x)} and k(x)=g(x)f(x)k(x) = \frac{g(x)}{f(x)}. Assuming f(a)≠0f(a) \neq 0, g(a)≠0g(a) \neq 0, and h′(a)h'(a) exists, which of the following correctly expresses k′(a)k'(a) in terms of h(a)h(a) and h′(a)h'(a)?
    Answer: k′(a)=−h′(a)[h(a)]2k'(a) = -\frac{h'(a)}{[h(a)]^2} Since k(x)=1h(x)k(x) = \frac{1}{h(x)}, we can find k′(x)k'(x) using the quotient rule (or reciprocal rule). Let u(x)=1u(x)=1 and v(x)=h(x)v(x)=h(x). Then k′(x)=u′v−uv′v2=(0)h(x)−(1)h′(x)[h(x)]2=−h′(x)[h(x)]2k'(x) = \frac{u'v - uv'}{v^2} = \frac{(0)h(x) - (1)h'(x)}{[h(x)]^2} = -\frac{h'(x)}{[h(x)]^2}. Evaluating at x=ax=a gives k′(a)=−h′(a)[h(a)]2k'(a) = -\frac{h'(a)}{[h(a)]^2}.
  5. The concentration CC of a medication in a patient's bloodstream tt hours after injection is modeled by C(t)=3tt2+9C(t) = \frac{3t}{t^2+9} for t≥0t \ge 0. At what time tt is the concentration of the medication increasing most rapidly?
    Answer: t=33t=3\sqrt{3} The rate of change of concentration is C′(t)C'(t). To find when this rate is at a maximum, we need to find the derivative of C′(t)C'(t), which is C′′(t)C''(t), and find its critical points. First, find C′(t)C'(t) using the Quotient Rule: C′(t)=(t2+9)(3)−(3t)(2t)(t2+9)2=3t2+27−6t2(t2+9)2=27−3t2(t2+9)2C'(t) = \frac{(t^2+9)(3) - (3t)(2t)}{(t^2+9)^2} = \frac{3t^2+27-6t^2}{(t^2+9)^2} = \frac{27-3t^2}{(t^2+9)^2} This is the rate of change. To maximize it, we find its derivative, C′′(t)C''(t), using the Quotient Rule again: C′′(t)=(−6t)(t2+9)2−(27−3t2)(2(t2+9)(2t))((t2+9)2)2C''(t) = \frac{(-6t)(t^2+9)^2 - (27-3t^2)(2(t^2+9)(2t))}{((t^2+9)^2)^2} Set C′′(t)=0C''(t)=0. This occurs when the numerator is 0. We can factor out −6t(t2+9)-6t(t^2+9) to simplify: Numerator = (t2+9)[(−6t)(t2+9)−4t(27−3t2)]=0(t^2+9)[(-6t)(t^2+9) - 4t(27-3t^2)] = 0 (−6t3−54t)−(108t−12t3)=0(-6t^3 - 54t) - (108t - 12t^3) = 0 6t3−162t=06t^3 - 162t = 0 6t(t2−27)=06t(t^2 - 27) = 0 This gives t=0t=0 or t2=27t^2=27. For t>0t>0, we have t=27=33t = \sqrt{27} = 3\sqrt{3}. This corresponds to an inflection point on the graph of C(t)C(t), where the rate of increase is maximal.
  6. If f(x)=xx−3f(x) = \frac{x}{x-3}, then what is the expression for the second derivative, f′′(x)f''(x)?
    Answer: 6(x−3)3\frac{6}{(x-3)^3} First, find the first derivative using the quotient rule: f′(x)=(1)(x−3)−(x)(1)(x−3)2=x−3−x(x−3)2=−3(x−3)2f'(x) = \frac{(1)(x-3) - (x)(1)}{(x-3)^2} = \frac{x-3-x}{(x-3)^2} = \frac{-3}{(x-3)^2}. To find the second derivative, we can write f′(x)=−3(x−3)−2f'(x) = -3(x-3)^{-2} and use the power rule and chain rule: f′′(x)=−3(−2)(x−3)−3(1)=6(x−3)−3=6(x−3)3f''(x) = -3(-2)(x-3)^{-3}(1) = 6(x-3)^{-3} = \frac{6}{(x-3)^3}. Alternatively, applying the quotient rule to f′(x)f'(x) gives f′′(x)=(0)(x−3)2−(−3)(2(x−3))(x−3)4=6(x−3)(x−3)4=6(x−3)3f''(x) = \frac{(0)(x-3)^2 - (-3)(2(x-3))}{(x-3)^4} = \frac{6(x-3)}{(x-3)^4} = \frac{6}{(x-3)^3}.
  7. Let f(x)=2x+3x+1f(x) = \frac{2x+3}{x+1} and let g(x)g(x) be the inverse function of f(x)f(x). What is the value of g′(1)g'(1)?
    Answer: −1-1 We use the formula for the derivative of an inverse function: g′(y0)=1f′(x0)g'(y_0) = \frac{1}{f'(x_0)}, where f(x0)=y0f(x_0) = y_0. Here, y0=1y_0=1. First, we find x0x_0 by solving f(x0)=1f(x_0) = 1: 2x0+3x0+1=1  ⟹  2x0+3=x0+1  ⟹  x0=−2\frac{2x_0+3}{x_0+1} = 1 \implies 2x_0+3 = x_0+1 \implies x_0 = -2. So, g(1)=−2g(1) = -2. Next, we find f′(x)f'(x) using the quotient rule: f′(x)=2(x+1)−(2x+3)(1)(x+1)2=2x+2−2x−3(x+1)2=−1(x+1)2f'(x) = \frac{2(x+1) - (2x+3)(1)}{(x+1)^2} = \frac{2x+2-2x-3}{(x+1)^2} = \frac{-1}{(x+1)^2}. Now we evaluate f′(−2)=−1(−2+1)2=−1(−1)2=−1f'(-2) = \frac{-1}{(-2+1)^2} = \frac{-1}{(-1)^2} = -1. Finally, g′(1)=1f′(−2)=1−1=−1g'(1) = \frac{1}{f'(-2)} = \frac{1}{-1} = -1.
  8. Let h(x)=u(x)1+u(x)h(x) = \frac{u(x)}{1+u(x)}. If u(2)=1u(2) = 1 and u′(2)=4u'(2) = 4, what is the value of h′(2)h'(2)?
    Answer: 11 We use the quotient rule to find h′(x)h'(x). The derivative of the numerator is u′(x)u'(x) and the derivative of the denominator is also u′(x)u'(x). h′(x)=u′(x)(1+u(x))−u(x)(u′(x))(1+u(x))2h'(x) = \frac{u'(x)(1+u(x)) - u(x)(u'(x))}{(1+u(x))^2}. Expanding the numerator: h′(x)=u′(x)+u′(x)u(x)−u(x)u′(x)(1+u(x))2=u′(x)(1+u(x))2h'(x) = \frac{u'(x) + u'(x)u(x) - u(x)u'(x)}{(1+u(x))^2} = \frac{u'(x)}{(1+u(x))^2}. Now, we substitute the given values at x=2x=2: h′(2)=u′(2)(1+u(2))2=4(1+1)2=422=44=1h'(2) = \frac{u'(2)}{(1+u(2))^2} = \frac{4}{(1+1)^2} = \frac{4}{2^2} = \frac{4}{4} = 1.
  9. Let f(x)=x3−8x2−4f(x) = \frac{x^3-8}{x^2-4} for x≠2x \neq 2. What is the slope of the line tangent to the graph of ff at x=1x=1?
    Answer: 59\frac{5}{9} When you need to find the slope of a tangent line at a specific point, you're looking for the derivative of the function evaluated at that point. The slope of the tangent line to f(x)f(x) at x=1x = 1 is simply f′(1)f'(1). To find f′(x)f'(x) for f(x)=x3−8x2−4f(x) = \frac{x^3-8}{x^2-4}, use the quotient rule: (uv)′=u′v−uv′v2\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}. Here, u=x3−8u = x^3-8 and v=x2−4v = x^2-4, so u′=3x2u' = 3x^2 and v′=2xv' = 2x. Applying the quotient rule: f′(x)=(3x2)(x2−4)−(x3−8)(2x)(x2−4)2f'(x) = \frac{(3x^2)(x^2-4) - (x^3-8)(2x)}{(x^2-4)^2} Expanding the numerator: f′(x)=3x4−12x2−2x4+16x(x2−4)2=x4−12x2+16x(x2−4)2f'(x) = \frac{3x^4 - 12x^2 - 2x^4 + 16x}{(x^2-4)^2} = \frac{x^4 - 12x^2 + 16x}{(x^2-4)^2} Now evaluate at x=1x = 1: f′(1)=1−12+16(1−4)2=59f'(1) = \frac{1 - 12 + 16}{(1-4)^2} = \frac{5}{9} This confirms answer D is correct. Answer A (73\frac{7}{3}) likely comes from arithmetic errors in the quotient rule application. Answer B (−59-\frac{5}{9}) gets the magnitude right but has the wrong sign, possibly from sign errors in the calculation. Answer C (11) might result from incorrectly simplifying the original function before differentiating or making computational mistakes. Remember: always double-check your quotient rule algebra, especially the signs. Expanding systematically and combining like terms carefully will help you avoid common arithmetic pitfalls in derivative problems.
  10. Let h(x)=f(x)g(x)h(x) = \frac{f(x)}{g(x)}. Suppose f(a)=4f(a)=4, f′(a)=−2f'(a)=-2, g(a)=1g(a)=1, and h′(a)=6h'(a)=6. What is the value of g′(a)g'(a)?
    Answer: −2-2 The Quotient Rule states that h′(x)=g(x)f′(x)−f(x)g′(x)[g(x)]2h'(x) = \frac{g(x)f'(x) - f(x)g'(x)}{[g(x)]^2}. We are given values at x=ax=a. Let's substitute them into the formula: h′(a)=g(a)f′(a)−f(a)g′(a)[g(a)]2h'(a) = \frac{g(a)f'(a) - f(a)g'(a)}{[g(a)]^2} Substitute the known values: h′(a)=6h'(a)=6, f(a)=4f(a)=4, f′(a)=−2f'(a)=-2, and g(a)=1g(a)=1. 6=(1)(−2)−(4)g′(a)126 = \frac{(1)(-2) - (4)g'(a)}{1^2} 6=−2−4g′(a)6 = -2 - 4g'(a) Now, we solve for g′(a)g'(a): 8=−4g′(a)8 = -4g'(a) g′(a)=8−4=−2g'(a) = \frac{8}{-4} = -2
  11. Let h(x)=f(x)g(x)h(x) = \frac{f(x)}{g(x)}. Given that f(2)=3f(2)=3, f′(2)=5f'(2)=5, g(2)=0g(2)=0, and g′(2)=4g'(2)=4, which of the following statements about the graph of h(x)h(x) at x=2x=2 is true?
    Answer: The graph has a vertical asymptote. The function h(x)=f(x)g(x)h(x) = \frac{f(x)}{g(x)} is defined where g(x)≠0g(x) \neq 0. Since we are given g(2)=0g(2)=0, the function h(x)h(x) is not defined at x=2x=2. This means there is a discontinuity at x=2x=2. To determine the type of discontinuity, we consider the limit of h(x)h(x) as xx approaches 2. The numerator approaches f(2)=3f(2)=3, which is a non-zero number. The denominator approaches g(2)=0g(2)=0. When the numerator of a rational function approaches a non-zero constant and the denominator approaches zero, the function has a vertical asymptote. Therefore, the graph of h(x)h(x) has a vertical asymptote at x=2x=2. A removable discontinuity would occur if both f(2)f(2) and g(2)g(2) were 0.
  12. Let differentiable functions ff and gg be given such that f(2)=3f(2) = 3, f′(2)=−1f'(2) = -1, g(2)=5g(2) = 5, and g′(2)=2g'(2) = 2. If H(x)=g(x)f(x)H(x) = \frac{g(x)}{f(x)}, what is the value of H′(2)H'(2)?
    Answer: 119\frac{11}{9} To find H′(x)H'(x), we apply the quotient rule: H′(x)=g′(x)f(x)−g(x)f′(x)[f(x)]2H'(x) = \frac{g'(x)f(x) - g(x)f'(x)}{[f(x)]^2}. Substituting the given values at x=2x=2: H′(2)=g′(2)f(2)−g(2)f′(2)[f(2)]2=(2)(3)−(5)(−1)32=6+59=119H'(2) = \frac{g'(2)f(2) - g(2)f'(2)}{[f(2)]^2} = \frac{(2)(3) - (5)(-1)}{3^2} = \frac{6 + 5}{9} = \frac{11}{9}.
  13. Given the information in the passage, what is the derivative of h(x)=f(x)g(x)h(x) = \frac{f(x)}{g(x)} at x=3x=3?
    Answer: 1316\frac{13}{16} From the tangent line information, we can deduce the function and derivative values at x=3x=3. For f(x)f(x), the point of tangency is f(3)=2(3)−1=5f(3) = 2(3)-1 = 5 and the slope is f′(3)=2f'(3) = 2. For g(x)g(x), the point of tangency is g(3)=−3+7=4g(3) = -3+7 = 4 and the slope is g′(3)=−1g'(3) = -1. Now, apply the quotient rule to h(x)h(x): h′(x)=f′(x)g(x)−f(x)g′(x)[g(x)]2h'(x) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}. At x=3x=3, h′(3)=f′(3)g(3)−f(3)g′(3)[g(3)]2=(2)(4)−(5)(−1)42=8+516=1316h'(3) = \frac{f'(3)g(3) - f(3)g'(3)}{[g(3)]^2} = \frac{(2)(4) - (5)(-1)}{4^2} = \frac{8+5}{16} = \frac{13}{16}.
  14. Let h(x)=cos⁡(x)x2+1h(x) = \frac{\cos(x)}{x^2+1}. What is the value of h′(π)h'(\pi)?
    Answer: 2π(π2+1)2\frac{2\pi}{(\pi^2+1)^2} To find h′(x)h'(x), we apply the Quotient Rule, (fg)′=f′g−fg′g2(\frac{f}{g})' = \frac{f'g - fg'}{g^2}, with f(x)=cos⁡(x)f(x) = \cos(x) and g(x)=x2+1g(x) = x^2+1. We have f′(x)=−sin⁡(x)f'(x) = -\sin(x) and g′(x)=2xg'(x) = 2x. h′(x)=(−sin⁡(x))(x2+1)−(cos⁡(x))(2x)(x2+1)2h'(x) = \frac{(-\sin(x))(x^2+1) - (\cos(x))(2x)}{(x^2+1)^2} Now, we evaluate h′(x)h'(x) at x=πx=\pi. We use the values sin⁡(π)=0\sin(\pi) = 0 and cos⁡(π)=−1\cos(\pi) = -1. h′(π)=(−sin⁡(π))(π2+1)−(cos⁡(π))(2π)(π2+1)2=(−0)(π2+1)−(−1)(2π)(π2+1)2=0+2π(π2+1)2=2π(π2+1)2h'(\pi) = \frac{(-\sin(\pi))(\pi^2+1) - (\cos(\pi))(2\pi)}{(\pi^2+1)^2} = \frac{(-0)(\pi^2+1) - (-1)(2\pi)}{(\pi^2+1)^2} = \frac{0 + 2\pi}{(\pi^2+1)^2} = \frac{2\pi}{(\pi^2+1)^2}
  15. If f(x)=sin⁡(πx)x2−1f(x) = \frac{\sin(\pi x)}{x^2-1}, what is f′(0)f'(0)?
    Answer: −π-\pi We use the quotient rule and the chain rule: f′(x)=(cos⁡(πx)⋅π)(x2−1)−(sin⁡(πx))(2x)(x2−1)2f'(x) = \frac{(\cos(\pi x) \cdot \pi)(x^2-1) - (\sin(\pi x))(2x)}{(x^2-1)^2}. To find f′(0)f'(0), we substitute x=0x=0 into the expression for the derivative: f′(0)=(cos⁡(0)⋅π)(02−1)−(sin⁡(0))(2⋅0)(02−1)2f'(0) = \frac{(\cos(0) \cdot \pi)(0^2-1) - (\sin(0))(2 \cdot 0)}{(0^2-1)^2}. Since cos⁡(0)=1\cos(0) = 1 and sin⁡(0)=0\sin(0) = 0, this simplifies to: f′(0)=(1⋅π)(−1)−(0)(0)(−1)2=−π1=−πf'(0) = \frac{(1 \cdot \pi)(-1) - (0)(0)}{(-1)^2} = \frac{-\pi}{1} = -\pi.
  16. If f(x)=x2x−2f(x) = \frac{x^2}{x-2}, then what is f′′(x)f''(x)?
    Answer: 8(x−2)3\frac{8}{(x-2)^3} First, we find the first derivative, f′(x)f'(x), using the Quotient Rule. Let u=x2u=x^2 and v=x−2v=x-2. Then u′=2xu'=2x and v′=1v'=1. f′(x)=(x−2)(2x)−(x2)(1)(x−2)2=2x2−4x−x2(x−2)2=x2−4x(x−2)2f'(x) = \frac{(x-2)(2x) - (x^2)(1)}{(x-2)^2} = \frac{2x^2 - 4x - x^2}{(x-2)^2} = \frac{x^2 - 4x}{(x-2)^2} Next, we find the second derivative, f′′(x)f''(x), by differentiating f′(x)f'(x) using the Quotient Rule again. Let u=x2−4xu = x^2 - 4x and v=(x−2)2v = (x-2)^2. Then u′=2x−4u' = 2x - 4 and v′=2(x−2)v' = 2(x-2). f′′(x)=((x−2)2)(2x−4)−(x2−4x)(2(x−2))((x−2)2)2f''(x) = \frac{((x-2)^2)(2x-4) - (x^2-4x)(2(x-2))}{((x-2)^2)^2} Factor out a common term (x−2)(x-2) from the numerator: f′′(x)=(x−2)[(x−2)(2x−4)−2(x2−4x)](x−2)4f''(x) = \frac{(x-2)[(x-2)(2x-4) - 2(x^2-4x)]}{(x-2)^4} f′′(x)=(2x2−4x−4x+8)−(2x2−8x)(x−2)3f''(x) = \frac{(2x^2 - 4x - 4x + 8) - (2x^2 - 8x)}{(x-2)^3} f′′(x)=2x2−8x+8−2x2+8x(x−2)3=8(x−2)3f''(x) = \frac{2x^2 - 8x + 8 - 2x^2 + 8x}{(x-2)^3} = \frac{8}{(x-2)^3}
  17. Let f(x)=x2−kx+kf(x) = \frac{x^2-k}{x+k}, where kk is a non-zero constant. The tangent line to the graph of ff at x=0x=0 is parallel to the line y=−x+5y = -x+5. What is the value of kk?
    Answer: −1-1 The slope of the line y=−x+5y=-x+5 is -1. For the tangent line to be parallel, its slope must also be -1. The slope of the tangent line is given by the derivative f′(x)f'(x). Using the quotient rule: f′(x)=(2x)(x+k)−(x2−k)(1)(x+k)2=2x2+2kx−x2+k(x+k)2=x2+2kx+k(x+k)2f'(x) = \frac{(2x)(x+k) - (x^2-k)(1)}{(x+k)^2} = \frac{2x^2+2kx-x^2+k}{(x+k)^2} = \frac{x^2+2kx+k}{(x+k)^2}. We need the slope at x=0x=0 to be -1. So, we evaluate f′(0)f'(0): f′(0)=02+2k(0)+k(0+k)2=kk2=1kf'(0) = \frac{0^2+2k(0)+k}{(0+k)^2} = \frac{k}{k^2} = \frac{1}{k}. Setting the slope equal to -1: 1k=−1\frac{1}{k} = -1, which implies k=−1k = -1.
  18. Let g(x)=lim⁡h→0cos⁡(x+h)−cos⁡(x)hg(x) = \lim_{h \to 0} \frac{\cos(x+h) - \cos(x)}{h}. What is the derivative of the function f(x)=g(x)xf(x) = \frac{g(x)}{x}?
    Answer: sin⁡(x)−xcos⁡(x)x2\frac{\sin(x) - x\cos(x)}{x^2} First, recognize that g(x)g(x) is the definition of the derivative of cos⁡(x)\cos(x). Therefore, g(x)=−sin⁡(x)g(x) = -\sin(x). The problem asks for the derivative of f(x)=−sin⁡(x)xf(x) = \frac{-\sin(x)}{x}. Using the quotient rule: f′(x)=(−cos⁡(x))(x)−(−sin⁡(x))(1)x2=−xcos⁡(x)+sin⁡(x)x2=sin⁡(x)−xcos⁡(x)x2f'(x) = \frac{(-\cos(x))(x) - (-\sin(x))(1)}{x^2} = \frac{-x\cos(x) + \sin(x)}{x^2} = \frac{\sin(x) - x\cos(x)}{x^2}.
  19. For what non-zero value of xx does the function f(x)=exx2f(x) = \frac{e^x}{x^2} have a horizontal tangent line?
    Answer: x=2x = 2 A horizontal tangent line occurs where the derivative f′(x)f'(x) is equal to zero. To find the derivative, we use the Quotient Rule: f′(x)=g(x)h′(x)−h(x)g′(x)[g(x)]2f'(x) = \frac{g(x)h'(x) - h(x)g'(x)}{[g(x)]^2}, where h(x)=exh(x) = e^x and g(x)=x2g(x) = x^2. The derivatives are h′(x)=exh'(x) = e^x and g′(x)=2xg'(x) = 2x. Applying the rule: f′(x)=x2⋅ex−ex⋅2x(x2)2=xex(x−2)x4=ex(x−2)x3f'(x) = \frac{x^2 \cdot e^x - e^x \cdot 2x}{(x^2)^2} = \frac{xe^x(x - 2)}{x^4} = \frac{e^x(x - 2)}{x^3} To find where the tangent line is horizontal, we set the derivative equal to zero: ex(x−2)x3=0\frac{e^x(x - 2)}{x^3} = 0 This occurs when the numerator is zero. Since exe^x is never zero, we must have x−2=0x - 2 = 0, which gives x=2x = 2.
  20. Let h(x)=f(x2)xh(x) = \frac{f(x^2)}{x}. Given that f(1)=3f(1) = 3 and f′(1)=−2f'(1) = -2, find h′(1)h'(1).
    Answer: −7-7 We use the quotient rule and the chain rule. The derivative of the numerator is ddxf(x2)=f′(x2)⋅2x\frac{d}{dx}f(x^2) = f'(x^2) \cdot 2x. Applying the quotient rule: h′(x)=(f′(x2)⋅2x)(x)−(f(x2))(1)x2h'(x) = \frac{(f'(x^2) \cdot 2x)(x) - (f(x^2))(1)}{x^2}. Now, substitute x=1x=1: h′(1)=(f′(12)⋅2(1))(1)−f(12)12=f′(1)⋅2−f(1)1=2f′(1)−f(1)h'(1) = \frac{(f'(1^2) \cdot 2(1))(1) - f(1^2)}{1^2} = \frac{f'(1) \cdot 2 - f(1)}{1} = 2f'(1) - f(1). Using the given values, h′(1)=2(−2)−3=−4−3=−7h'(1) = 2(-2) - 3 = -4 - 3 = -7.

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