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Derivatives of tan & Other Trig - Calculus 1

  1. What is the derivative of f(x)=tan⁡(2x)x2+1f(x) = \frac{\tan(2x)}{x^2+1}?
    Answer: 2sec⁡2(2x)(x2+1)−2xtan⁡(2x)(x2+1)2\frac{2\sec^2(2x)(x^2+1) - 2x\tan(2x)}{(x^2+1)^2} Use the quotient rule (uv)′=u′v−uv′v2(\frac{u}{v})' = \frac{u'v - uv'}{v^2} with u=tan⁡(2x)u = \tan(2x) and $ v = x2x^2+1 $. First, find the derivatives: $ u' = \sec^2(2x) \cdot 2 = 2\sec^2(2x) $ (by the chain rule) and $ v' = 2x $. Now, substitute into the quotient rule formula: f′(x)=(2sec⁡2(2x))(x2+1)−(tan⁡(2x))(2x)(x2+1)2=2sec⁡2(2x)(x2+1)−2xtan⁡(2x)(x2+1)2f'(x) = \frac{(2\sec^2(2x))(x^2+1) - (\tan(2x))(2x)}{(x^2+1)^2} = \frac{2\sec^2(2x)(x^2+1) - 2x\tan(2x)}{(x^2+1)^2}.
  2. Find the equation of the line normal to the graph of f(x)=sec⁡(x)f(x) = \sec(x) at the point where x=π/3x = \pi/3.
    Answer: y−2=−123(x−π3)y - 2 = -\frac{1}{2\sqrt{3}}(x - \frac{\pi}{3}) The normal line is perpendicular to the tangent line. First, find the slope of the tangent line, $ mtanm_{tan} $, by evaluating the derivative at x=π/3x = \pi/3. f′(x)=sec⁡(x)tan⁡(x)f'(x) = \sec(x)\tan(x). At x=π/3x=\pi/3, $ mtanm_{tan} = \sec(π/3\pi/3)\tan(π/3\pi/3) = (2)(3\sqrt{3}) = 2\sqrt{3} $. The slope of the normal line, $ mnormm_{norm} $, is the negative reciprocal of the tangent slope: $ mnormm_{norm} = -\frac{1}{m_{tan}} = -\frac{1}{2\sqrt{3}} $. The point of interest is (π/3,f(π/3))=(π/3,sec⁡(π/3))=(π/3,2)(\pi/3, f(\pi/3)) = (\pi/3, \sec(\pi/3)) = (\pi/3, 2). Using the point-slope form, the equation of the normal line is y−2=−123(x−π3)y - 2 = -\frac{1}{2\sqrt{3}}(x - \frac{\pi}{3}).
  3. Let f(x)=sec⁡(x)1+tan⁡(x)f(x) = \frac{\sec(x)}{1+\tan(x)}. Find the value of f′(π)f'(\pi).
    Answer: 11 When you need to find the derivative of a function involving trigonometric expressions, start by simplifying the function if possible before differentiating. This often makes the calculation much more manageable. For f(x)=sec⁡(x)1+tan⁡(x)f(x) = \frac{\sec(x)}{1+\tan(x)}, let's first rewrite this in terms of sine and cosine: f(x)=1cos⁡(x)1+sin⁡(x)cos⁡(x)=1cos⁡(x)cos⁡(x)+sin⁡(x)cos⁡(x)=1cos⁡(x)+sin⁡(x)f(x) = \frac{\frac{1}{\cos(x)}}{1+\frac{\sin(x)}{\cos(x)}} = \frac{\frac{1}{\cos(x)}}{\frac{\cos(x)+\sin(x)}{\cos(x)}} = \frac{1}{\cos(x)+\sin(x)} Now we can differentiate: f′(x)=−(−sin⁡(x)+cos⁡(x))(cos⁡(x)+sin⁡(x))2=sin⁡(x)−cos⁡(x)(cos⁡(x)+sin⁡(x))2f'(x) = \frac{-(-\sin(x)+\cos(x))}{(\cos(x)+\sin(x))^2} = \frac{\sin(x)-\cos(x)}{(\cos(x)+\sin(x))^2} At x=πx = \pi: sin⁡(π)=0\sin(\pi) = 0 and cos⁡(π)=−1\cos(\pi) = -1, so: f′(π)=0−(−1)((−1)+0)2=11=1f'(\pi) = \frac{0-(-1)}{((-1)+0)^2} = \frac{1}{1} = 1 Answer C) 11 is correct. Answer A) 1/21/2 likely results from errors in the quotient rule application or trigonometric value substitution. Answer B) −1-1 could come from sign errors when evaluating sin⁡(π)−cos⁡(π)\sin(\pi) - \cos(\pi) or forgetting that cos⁡(π)=−1\cos(\pi) = -1. Answer D) "Undefined" might seem tempting since the original function appears to have issues when tan⁡(x)=−1\tan(x) = -1, but at x=πx = \pi, both the function and its derivative are well-defined. Study tip: Always simplify trigonometric expressions to basic sine and cosine forms before differentiating. This prevents errors and often reveals that apparently complex derivatives are actually straightforward.
  4. Evaluate the limit: lim⁡x→π/2tan⁡(2x)cot⁡(x)\lim_{x \to \pi/2} \frac{\tan(2x)}{\cot(x)}
    Answer: −2-2 When you encounter limits involving trigonometric functions that create indeterminate forms, the key strategy is to rewrite the expression using fundamental trigonometric identities before applying limit properties. Let's rewrite this expression using basic trig identities. Since tan⁡(2x)=sin⁡(2x)cos⁡(2x)\tan(2x) = \frac{\sin(2x)}{\cos(2x)} and cot⁡(x)=cos⁡(x)sin⁡(x)\cot(x) = \frac{\cos(x)}{\sin(x)}, we have: tan⁡(2x)cot⁡(x)=sin⁡(2x)cos⁡(2x)⋅sin⁡(x)cos⁡(x)=sin⁡(2x)sin⁡(x)cos⁡(2x)cos⁡(x)\frac{\tan(2x)}{\cot(x)} = \frac{\sin(2x)}{\cos(2x)} \cdot \frac{\sin(x)}{\cos(x)} = \frac{\sin(2x)\sin(x)}{\cos(2x)\cos(x)} Using the double angle identity sin⁡(2x)=2sin⁡(x)cos⁡(x)\sin(2x) = 2\sin(x)\cos(x): sin⁡(2x)sin⁡(x)cos⁡(2x)cos⁡(x)=2sin⁡(x)cos⁡(x)sin⁡(x)cos⁡(2x)cos⁡(x)=2sin⁡2(x)cos⁡(2x)\frac{\sin(2x)\sin(x)}{\cos(2x)\cos(x)} = \frac{2\sin(x)\cos(x)\sin(x)}{\cos(2x)\cos(x)} = \frac{2\sin^2(x)}{\cos(2x)} Since cos⁡(2x)=cos⁡2(x)−sin⁡2(x)=1−2sin⁡2(x)\cos(2x) = \cos^2(x) - \sin^2(x) = 1 - 2\sin^2(x), we get: 2sin⁡2(x)1−2sin⁡2(x)\frac{2\sin^2(x)}{1 - 2\sin^2(x)} As x→π/2x \to \pi/2, we have sin⁡(x)→1\sin(x) \to 1, so sin⁡2(x)→1\sin^2(x) \to 1. Therefore: lim⁡x→π/22sin⁡2(x)1−2sin⁡2(x)=2(1)1−2(1)=2−1=−2\lim_{x \to \pi/2} \frac{2\sin^2(x)}{1 - 2\sin^2(x)} = \frac{2(1)}{1 - 2(1)} = \frac{2}{-1} = -2 Choice A (2) ignores the negative sign in the denominator. Choice C (1) likely comes from incorrectly simplifying the trigonometric identities. Choice D (does not exist) assumes the limit is undefined, but our algebraic manipulation shows it exists and equals -2. Study tip: When limits involve multiple trig functions, always convert to sines and cosines first, then apply double angle formulas to simplify before substituting the limit value.
  5. Find the x-coordinates in the interval (0,π)(0, \pi) where the tangent line to f(x)=2tan⁡(x)f(x) = 2\tan(x) is parallel to the line $ y=4x-1 $.
    Answer: x=π/4x = \pi/4 and x=3π/4x = 3\pi/4 The tangent line is parallel to $ y=4x-1 $ if its slope is 4. The slope is given by the derivative $ f'(x) $. The derivative of f(x)=2tan⁡(x)f(x)=2\tan(x) is f′(x)=2sec⁡2(x)f'(x) = 2\sec^2(x). Set the derivative equal to 4: 2sec⁡2(x)=42\sec^2(x) = 4, which simplifies to sec⁡2(x)=2\sec^2(x) = 2. Taking the square root gives sec⁡(x)=±2\sec(x) = \pm\sqrt{2}, which is equivalent to cos⁡(x)=±12\cos(x) = \pm\frac{1}{\sqrt{2}}. In the interval (0,π)(0, \pi), the solutions are $ x = \frac{\pi}{4} $ (where cos⁡(x)=12\cos(x) = \frac{1}{\sqrt{2}}) and $ x = \frac{3\pi}{4} $ (where $ \cos(x) = -\frac{1}{\sqrt{2}} $).
  6. The height of a weather balloon is given by h(t)=50tan⁡(π12t)h(t) = 50 \tan(\frac{\pi}{12}t) for 0≤t<60 \le t < 6, where hh is in meters and tt is in hours. What is the balloon's vertical acceleration at t=2t=2 hours?
    Answer: 25π2381\frac{25\pi^2\sqrt{3}}{81} m/hr2^2 When you see a question asking for acceleration given a position function, remember that acceleration is the second derivative of position. You'll need to differentiate twice to get from height to vertical acceleration. Starting with h(t)=50tan⁡(π12t)h(t) = 50\tan\left(\frac{\pi}{12}t\right), let's find the velocity by taking the first derivative. Using the chain rule: h′(t)=50⋅sec⁡2(π12t)⋅π12=25π6sec⁡2(π12t)h'(t) = 50 \cdot \sec^2\left(\frac{\pi}{12}t\right) \cdot \frac{\pi}{12} = \frac{25\pi}{6}\sec^2\left(\frac{\pi}{12}t\right) For acceleration, we need the second derivative. The derivative of sec⁡2(u)\sec^2(u) is 2sec⁡(u)⋅sec⁡(u)tan⁡(u)=2sec⁡2(u)tan⁡(u)2\sec(u) \cdot \sec(u)\tan(u) = 2\sec^2(u)\tan(u). Applying the chain rule again: h′′(t)=25π6⋅2sec⁡2(π12t)tan⁡(π12t)⋅π12=25π236sec⁡2(π12t)tan⁡(π12t)h''(t) = \frac{25\pi}{6} \cdot 2\sec^2\left(\frac{\pi}{12}t\right)\tan\left(\frac{\pi}{12}t\right) \cdot \frac{\pi}{12} = \frac{25\pi^2}{36}\sec^2\left(\frac{\pi}{12}t\right)\tan\left(\frac{\pi}{12}t\right) At t=2t = 2: π12⋅2=π6\frac{\pi}{12} \cdot 2 = \frac{\pi}{6}, so we need sec⁡2(π6)tan⁡(π6)\sec^2\left(\frac{\pi}{6}\right)\tan\left(\frac{\pi}{6}\right). Since cos⁡(π6)=32\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} and tan⁡(π6)=13\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}, we get sec⁡2(π6)=43\sec^2\left(\frac{\pi}{6}\right) = \frac{4}{3}. Therefore: h′′(2)=25π236⋅43⋅13=25π2381h''(2) = \frac{25\pi^2}{36} \cdot \frac{4}{3} \cdot \frac{1}{\sqrt{3}} = \frac{25\pi^2\sqrt{3}}{81}, which is choice C. Choice A uses the wrong coefficient. Choice B gives the first derivative evaluation, not the second. Choice D forgets to evaluate the trigonometric functions at t=2t=2. Remember: acceleration problems always require the second derivative, and don't forget to evaluate trigonometric expressions at the given point.
  7. Let f(x)=exsec⁡(x)f(x) = e^x \sec(x) for xx in the interval (−π/2,π/2)(-\pi/2, \pi/2). Find the xx-coordinate of the point where the tangent line to the graph of ff is horizontal.
    Answer: x=−π/4x = -\pi/4 A horizontal tangent line occurs where the derivative is zero. We need to find f′(x)f'(x) using the product rule: f′(x)=ddx(ex)sec⁡(x)+exddx(sec⁡(x))f'(x) = \frac{d}{dx}(e^x)\sec(x) + e^x \frac{d}{dx}(\sec(x)). This gives f′(x)=exsec⁡(x)+ex(sec⁡(x)tan⁡(x))f'(x) = e^x \sec(x) + e^x (\sec(x)\tan(x)). Set f′(x)=0f'(x) = 0: exsec⁡(x)(1+tan⁡(x))=0e^x \sec(x) (1 + \tan(x)) = 0. In the interval (−π/2,π/2)(-\pi/2, \pi/2), both exe^x and sec⁡(x)\sec(x) are always positive. Thus, for the product to be zero, we must have 1+tan⁡(x)=01 + \tan(x) = 0, which means tan⁡(x)=−1\tan(x) = -1. The solution to this equation in the interval (−π/2,π/2)(-\pi/2, \pi/2) is x=−π/4x = -\pi/4. - B is incorrect: This would be the solution if tan⁡(x)=1\tan(x) = 1, a sign error. - C is incorrect: At x=0x=0, f′(0)=e0sec⁡(0)(1+tan⁡(0))=1(1)(1+0)=1f'(0) = e^0 \sec(0) (1 + \tan(0)) = 1(1)(1+0) = 1, not 0. - D is incorrect: A solution exists, as shown above.
  8. If y=csc⁡(x)x2y = \frac{\csc(x)}{x^2}, then dydx=\frac{dy}{dx} =?
    Answer: −csc⁡(x)(xcot⁡(x)+2)x3\frac{-\csc(x)(x\cot(x) + 2)}{x^3} Using the quotient rule, (fg)′=f′g−fg′g2(\frac{f}{g})' = \frac{f'g - fg'}{g^2}, with f(x)=csc⁡(x)f(x) = \csc(x) and $ g(x) = x2x^2 $. We have f′(x)=−csc⁡(x)cot⁡(x)f'(x) = -\csc(x)\cot(x) and $ g'(x) = 2x $. Plugging these into the formula gives dydx=(−csc⁡(x)cot⁡(x))(x2)−(csc⁡(x))(2x)(x2)2=−x2csc⁡(x)cot⁡(x)−2xcsc⁡(x)x4\frac{dy}{dx} = \frac{(-\csc(x)\cot(x))(x^2) - (\csc(x))(2x)}{(x^2)^2} = \frac{-x^2\csc(x)\cot(x) - 2x\csc(x)}{x^4}. Factoring −xcsc⁡(x)-x\csc(x) from the numerator gives −xcsc⁡(x)(xcot⁡(x)+2)x4\frac{-x\csc(x)(x\cot(x) + 2)}{x^4}. Simplifying by canceling an $ x $ gives −csc⁡(x)(xcot⁡(x)+2)x3\frac{-\csc(x)(x\cot(x) + 2)}{x^3}.
  9. Let g(x)=cot⁡(x)g(x) = \cot(x). Find the second derivative, $ g''(x) $.
    Answer: 2csc⁡2(x)cot⁡(x)2\csc^2(x)\cot(x) First, find the first derivative: g′(x)=−csc⁡2(x)g'(x) = -\csc^2(x). To find the second derivative, we differentiate $ g'(x) $. This requires the chain rule. Let u=csc⁡(x)u = \csc(x), so $ g'(x) = -u2u^2 $. Then g′′(x)=−2u⋅u′g''(x) = -2u \cdot u'. The derivative of u=csc⁡(x)u = \csc(x) is u′=−csc⁡(x)cot⁡(x)u' = -\csc(x)\cot(x). Substituting back, we get g′′(x)=−2(csc⁡(x))(−csc⁡(x)cot⁡(x))=2csc⁡2(x)cot⁡(x)g''(x) = -2(\csc(x))(-\csc(x)\cot(x)) = 2\csc^2(x)\cot(x).
  10. If y=(1+tan⁡(x))3y = (1 + \tan(x))^3, find the slope of the tangent line at $ x=0 $.
    Answer: 3 The slope of the tangent line is the value of the derivative at that point. We use the chain rule. Let u=1+tan⁡(x)u = 1 + \tan(x), so $ y = u3u^3 $. Then dydu=3u2\frac{dy}{du} = 3u^2 and dudx=sec⁡2(x)\frac{du}{dx} = \sec^2(x). So, dydx=3u2⋅sec⁡2(x)=3(1+tan⁡(x))2sec⁡2(x)\frac{dy}{dx} = 3u^2 \cdot \sec^2(x) = 3(1 + \tan(x))^2 \sec^2(x). Now, evaluate at $ x=0 $. We have tan⁡(0)=0\tan(0) = 0 and sec⁡(0)=1\sec(0) = 1. Plugging these values in: slope=3(1+0)2(1)2=3(1)(1)=3\text{slope} = 3(1 + 0)^2 (1)^2 = 3(1)(1) = 3.
  11. Find the smallest positive value of $ x $ for which the derivative of f(x)=sec⁡(x)+csc⁡(x)f(x) = \sec(x) + \csc(x) is zero.
    Answer: π/4\pi/4 First, find the derivative: f′(x)=sec⁡(x)tan⁡(x)−csc⁡(x)cot⁡(x)f'(x) = \sec(x)\tan(x) - \csc(x)\cot(x). Set the derivative to zero: sec⁡(x)tan⁡(x)=csc⁡(x)cot⁡(x)\sec(x)\tan(x) = \csc(x)\cot(x). Rewrite in terms of sine and cosine: 1cos⁡(x)sin⁡(x)cos⁡(x)=1sin⁡(x)cos⁡(x)sin⁡(x)\frac{1}{\cos(x)} \frac{\sin(x)}{\cos(x)} = \frac{1}{\sin(x)} \frac{\cos(x)}{\sin(x)}, which simplifies to sin⁡(x)cos⁡2(x)=cos⁡(x)sin⁡2(x)\frac{\sin(x)}{\cos^2(x)} = \frac{\cos(x)}{\sin^2(x)}. Cross-multiply to get sin⁡3(x)=cos⁡3(x)\sin^3(x) = \cos^3(x). This implies tan⁡3(x)=1\tan^3(x) = 1, so tan⁡(x)=1\tan(x) = 1. The smallest positive value of $ x $ for which this is true is x=π/4x = \pi/4.
  12. The derivative of f(x)=tan⁡(x) f(x) = \tan(\sqrt{x}) is:
    Answer: sec⁡2(x)2x \frac{\sec^2(\sqrt{x})}{2\sqrt{x}} This problem requires the chain rule. Let the outer function be g(u)=tan⁡(u) g(u) = \tan(u) and the inner function be u(x)=x=x1/2 u(x) = \sqrt{x} = x^{1/2} . The derivatives are g′(u)=sec⁡2(u) g'(u) = \sec^2(u) and u′(x)=12x−1/2=12x u'(x) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}} . By the chain rule, f′(x)=g′(u(x))⋅u′(x)=sec⁡2(x)⋅12x=sec⁡2(x)2x f'(x) = g'(u(x)) \cdot u'(x) = \sec^2(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} = \frac{\sec^2(\sqrt{x})}{2\sqrt{x}} .
  13. If sec⁡(xy)=y\sec(xy) = y, find dydx\frac{dy}{dx} at the point (π,−1)(\pi, -1).
    Answer: $ 0 $ Differentiate both sides implicitly with respect to $ x $. The left side requires the chain rule and product rule: sec⁡(xy)tan⁡(xy)⋅(1⋅y+xdydx)=dydx\sec(xy)\tan(xy) \cdot (1\cdot y + x \frac{dy}{dx}) = \frac{dy}{dx}. Now, substitute the point x=π,y=−1x=\pi, y=-1. The argument of the trig functions is xy=−πxy = -\pi. sec⁡(−π)=−1\sec(-\pi) = -1 and tan⁡(−π)=0\tan(-\pi) = 0. Substituting these values into the differentiated equation gives: (−1)(0)⋅(−1+πdydx)=dydx(-1)(0) \cdot (-1 + \pi \frac{dy}{dx}) = \frac{dy}{dx}. This simplifies to 0=dydx0 = \frac{dy}{dx}.
  14. Let f(x)=ln⁡(sec⁡(x)+tan⁡(x)) f(x) = \ln(\sec(x) + \tan(x)) . What is f′(x) f'(x) ?
    Answer: sec⁡(x) \sec(x) Using the chain rule, f′(x)=1sec⁡(x)+tan⁡(x)⋅ddx(sec⁡(x)+tan⁡(x)) f'(x) = \frac{1}{\sec(x) + \tan(x)} \cdot \frac{d}{dx}(\sec(x) + \tan(x)) . The derivative of the inner part is sec⁡(x)tan⁡(x)+sec⁡2(x) \sec(x)\tan(x) + \sec^2(x) . So, f′(x)=sec⁡(x)tan⁡(x)+sec⁡2(x)sec⁡(x)+tan⁡(x) f'(x) = \frac{\sec(x)\tan(x) + \sec^2(x)}{\sec(x) + \tan(x)} . Factoring sec⁡(x) \sec(x) from the numerator gives f′(x)=sec⁡(x)(tan⁡(x)+sec⁡(x))sec⁡(x)+tan⁡(x) f'(x) = \frac{\sec(x)(\tan(x) + \sec(x))}{\sec(x) + \tan(x)} . The term (sec⁡(x)+tan⁡(x)) (\sec(x) + \tan(x)) cancels, leaving f′(x)=sec⁡(x) f'(x) = \sec(x) .
  15. The function f(x)=tan⁡(x)−x f(x) = \tan(x) - x is increasing on the interval (−π/2,π/2) (-\pi/2, \pi/2) when:
    Answer: f′(x)>0 f'(x) > 0 A function is increasing when its first derivative is positive. First, find the derivative: f′(x)=sec⁡2(x)−1 f'(x) = \sec^2(x) - 1 . Using the identity sec⁡2(x)−1=tan⁡2(x) \sec^2(x) - 1 = \tan^2(x) , we have f′(x)=tan⁡2(x) f'(x) = \tan^2(x) . Since the square of any real number is non-negative, tan⁡2(x)≥0 \tan^2(x) \ge 0 for all x x in the domain. The function is increasing wherever f′(x)>0 f'(x) > 0 . This is true for all x x in (−π/2,π/2) (-\pi/2, \pi/2) except at x=0 x=0 , where f′(0)=0 f'(0)=0 . So the condition for increasing is f′(x)>0 f'(x) > 0 .
  16. Evaluate the limit: lim⁡h→0sec⁡(π3+h)−sec⁡(π3)h \lim_{h \to 0} \frac{\sec(\frac{\pi}{3} + h) - \sec(\frac{\pi}{3})}{h}
    Answer: 23 2\sqrt{3} This limit represents the definition of the derivative of the function f(x)=sec⁡(x) f(x) = \sec(x) at the point x=π3 x = \frac{\pi}{3} . First, find the derivative of f(x) f(x) : f′(x)=sec⁡(x)tan⁡(x) f'(x) = \sec(x)\tan(x) . Then, evaluate this derivative at x=π3 x = \frac{\pi}{3} : f′(π3)=sec⁡(π3)tan⁡(π3)=(2)(3)=23 f'(\frac{\pi}{3}) = \sec(\frac{\pi}{3})\tan(\frac{\pi}{3}) = (2)(\sqrt{3}) = 2\sqrt{3} .
  17. If h(x)=sec⁡(tan⁡(x)) h(x) = \sec(\tan(x)) , find h′(x) h'(x) .
    Answer: sec⁡2(x)sec⁡(tan⁡(x))tan⁡(tan⁡(x)) \sec^2(x) \sec(\tan(x)) \tan(\tan(x)) This requires the chain rule. The outer function is f(u)=sec⁡(u) f(u) = \sec(u) and the inner function is u=tan⁡(x) u = \tan(x) . The derivative of the outer function is f′(u)=sec⁡(u)tan⁡(u) f'(u) = \sec(u)\tan(u) . The derivative of the inner function is u′=sec⁡2(x) u' = \sec^2(x) . By the chain rule, h′(x)=f′(u)⋅u′=sec⁡(tan⁡(x))tan⁡(tan⁡(x))⋅sec⁡2(x) h'(x) = f'(u) \cdot u' = \sec(\tan(x))\tan(\tan(x)) \cdot \sec^2(x) .
  18. Let h(x)=csc⁡(x)h(x) = \csc(x). Find the value of the second derivative, h′′(π/2)h''(\pi/2).
    Answer: 11 When you encounter second derivative problems involving trigonometric functions, you'll need to apply the chain rule and remember key trigonometric identities and derivatives. To find h′′(π/2)h''(\pi/2) where h(x)=csc⁡(x)h(x) = \csc(x), start by finding the first derivative. Since csc⁡(x)=1sin⁡(x)\csc(x) = \frac{1}{\sin(x)}, we have h′(x)=−csc⁡(x)cot⁡(x)h'(x) = -\csc(x)\cot(x). For the second derivative, differentiate h′(x)=−csc⁡(x)cot⁡(x)h'(x) = -\csc(x)\cot(x) using the product rule: h′′(x)=−[csc⁡′(x)cot⁡(x)+csc⁡(x)cot⁡′(x)]h''(x) = -[\csc'(x)\cot(x) + \csc(x)\cot'(x)] =−[−csc⁡(x)cot⁡(x)⋅cot⁡(x)+csc⁡(x)⋅(−csc⁡2(x))]= -[-\csc(x)\cot(x) \cdot \cot(x) + \csc(x) \cdot (-\csc^2(x))] =csc⁡(x)cot⁡2(x)+csc⁡3(x)= \csc(x)\cot^2(x) + \csc^3(x) =csc⁡(x)[cot⁡2(x)+csc⁡2(x)]= \csc(x)[\cot^2(x) + \csc^2(x)] Now evaluate at x=π/2x = \pi/2: csc⁡(π/2)=1\csc(\pi/2) = 1 and cot⁡(π/2)=0\cot(\pi/2) = 0. Therefore: h′′(π/2)=1[02+12]=1h''(\pi/2) = 1[0^2 + 1^2] = 1 Looking at the wrong answers: Choice (A) 2 likely comes from incorrectly applying the product rule or making an algebraic error. Choice (B) -1 probably results from a sign error during differentiation. Choice (C) 0 might occur if you mistakenly think that since cot⁡(π/2)=0\cot(\pi/2) = 0, the entire expression equals zero. Study tip: When differentiating products of trigonometric functions, write out each step of the product rule carefully and double-check your trigonometric values at special angles like π/2\pi/2.
  19. Let f(x)=ktan⁡(x)f(x) = k \tan(x), where $ k $ is a constant. If the slope of the tangent line to $ f(x) $ at x=π4x = \frac{\pi}{4} is 8, what is the value of $ k $?
    Answer: 4 The slope of the tangent line is given by the derivative $ f'(x) $. First, find the derivative: f′(x)=ksec⁡2(x)f'(x) = k \sec^2(x). The problem states that the slope at x=π4x = \frac{\pi}{4} is 8. So, f′(π4)=8f'(\frac{\pi}{4}) = 8. We have ksec⁡2(π4)=8k \sec^2(\frac{\pi}{4}) = 8. Since sec⁡(π4)=2\sec(\frac{\pi}{4}) = \sqrt{2}, we have sec⁡2(π4)=2\sec^2(\frac{\pi}{4}) = 2. The equation becomes $ k(2) = 8 $, which implies $ k=4 $.
  20. For which value of $ x $ in the interval (−π/2,π/2)(-\pi/2, \pi/2) is the tangent line to the graph of f(x)=tan⁡(x)−4xf(x) = \tan(x) - 4x horizontal?
    Answer: π3\frac{\pi}{3} A horizontal tangent line occurs when the derivative is zero. First, find the derivative: f′(x)=sec⁡2(x)−4f'(x) = \sec^2(x) - 4. Set the derivative to zero and solve for $ x $: sec⁡2(x)−4=0\sec^2(x) - 4 = 0, which means sec⁡2(x)=4\sec^2(x) = 4. Taking the square root gives sec⁡(x)=±2\sec(x) = \pm 2, which is equivalent to cos⁡(x)=±12\cos(x) = \pm \frac{1}{2}. In the interval (−π/2,π/2)(-\pi/2, \pi/2), cos⁡(x)\cos(x) is positive. So we need to solve cos⁡(x)=12\cos(x) = \frac{1}{2}. The solution in this interval is x=π3x = \frac{\pi}{3}.

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