Derivatives of tan, cot, sec, csc - Calculus 1

1. Let $f(x) = e^x \sec(x)$. What is the value of $f'(0)$?
   Answer: 1 To find the derivative of $f(x) = e^x \sec(x)$, we must use the product rule, $(uv)' = u'v + uv'$. Let $u = e^x$ and $v = \sec(x)$. Then $u' = e^x$ and $v' = \sec(x)\tan(x)$. Applying the product rule: $$f'(x) = (e^x)'\sec(x) + e^x(\sec(x))' = e^x\sec(x) + e^x\sec(x)\tan(x)$$ Now, we evaluate $f'(x)$ at $x=0$: $$f'(0) = e^0\sec(0) + e^0\sec(0)\tan(0)$$ Since $e^0=1$, $\sec(0)=1$, and $\tan(0)=0$, we have: $$f'(0) = (1)(1) + (1)(1)(0) = 1 + 0 = 1$$
2. Let $h(x) = \tan(x) \sec(2x)$. What is the value of $h'(0)$?
   Answer: 1 We use the product rule, $(uv)' = u'v + uv'$, where $u = \tan(x)$ and $v = \sec(2x)$. $u' = \sec^2(x)$. $v' = \sec(2x)\tan(2x) \cdot 2 = 2\sec(2x)\tan(2x)$ (by the chain rule). So, $h'(x) = \sec^2(x)\sec(2x) + \tan(x)[2\sec(2x)\tan(2x)]$. Now, evaluate at $x = 0$: $h'(0) = \sec^2(0)\sec(0) + \tan(0)[2\sec(0)\tan(0)]$. Since $\sec(0) = 1$ and $\tan(0) = 0$: $h'(0) = (1)^2(1) + (0)[2(1)(0)] = 1 + 0 = 1$. Distractor A results from thinking the $\tan(0)$ term makes the entire expression zero. Other distractors stem from errors in applying the chain or product rule.
3. What is an equation of the line tangent to the graph of $y = 2\csc(x)$ at the point where $x = \frac{5\pi}{6}$?
   Answer: $y - 4 = 4\sqrt{3}(x - \frac{5\pi}{6})$ When finding the equation of a tangent line to any function, you need two key pieces: a point on the line and the slope at that point. The slope comes from taking the derivative and evaluating it at the given x-value. First, let's find the y-coordinate when $x = \frac{5\pi}{6}$. Since $\sin(\frac{5\pi}{6}) = \frac{1}{2}$, we have $y = 2\csc(\frac{5\pi}{6}) = \frac{2}{\sin(\frac{5\pi}{6})} = \frac{2}{\frac{1}{2}} = 4$. So our point is $(\frac{5\pi}{6}, 4)$. Next, we need the slope by finding $\frac{d}{dx}[2\csc(x)]$. The derivative of $\csc(x)$ is $-\csc(x)\cot(x)$, so $\frac{d}{dx}[2\csc(x)] = -2\csc(x)\cot(x)$. At $x = \frac{5\pi}{6}$: $\csc(\frac{5\pi}{6}) = 2$ and $\cot(\frac{5\pi}{6}) = \frac{\cos(\frac{5\pi}{6})}{\sin(\frac{5\pi}{6})} = \frac{-\frac{\sqrt{3}}{2}}{\frac{1}{2}} = -\sqrt{3}$. Therefore, the slope is $-2(2)(-\sqrt{3}) = 4\sqrt{3}$. Using point-slope form with point $(\frac{5\pi}{6}, 4)$ and slope $4\sqrt{3}$: $y - 4 = 4\sqrt{3}(x - \frac{5\pi}{6})$, which is choice C. Choice A has the wrong y-coordinate in the point-slope form. Choice B uses the correct point but has the wrong sign on the slope. Choice D has the correct point but is missing the factor of 4 in the slope. Remember: for tangent line problems, always find both the point and the derivative value systematically—small sign errors in trigonometric derivatives are common pitfalls.
4. Which of the following is the value of the limit $\lim_{h \to 0} \frac{\sec(\pi/3 + h) - \sec(\pi/3)}{h}$?
   Answer: $2\sqrt{3}$ This limit represents the definition of the derivative of the function $f(x) = \sec(x)$ at the point $x = \pi/3$. The derivative of $f(x) = \sec(x)$ is $f'(x) = \sec(x)\tan(x)$. We need to evaluate this derivative at $x = \pi/3$: $f'(\pi/3) = \sec(\pi/3)\tan(\pi/3)$. We know that $\sec(\pi/3) = 2$ and $\tan(\pi/3) = \sqrt{3}$. Therefore, the value of the limit is $(2)(\sqrt{3}) = 2\sqrt{3}$. Distractor A is the value of the function $\sec(\pi/3)$. Distractor B is the value of $\sec^2(\pi/3)$, which would be the derivative of $\tan(x)$ at that point.
5. If $y = \frac{\csc(x)}{1 + \cot(x)}$, which of the following is an expression for $\frac{dy}{dx}$?
   Answer: $\frac{\sin(x) - \cos(x)}{(\sin(x) + \cos(x))^2}$ While the quotient rule can be used, it's easier to first simplify the expression using trigonometric identities: $y = \frac{1/\sin(x)}{1 + \cos(x)/\sin(x)} = \frac{1/\sin(x)}{(\sin(x) + \cos(x))/\sin(x)} = \frac{1}{\sin(x) + \cos(x)}$. Now, we can write $y = (\sin(x) + \cos(x))^{-1}$ and use the chain rule: $\frac{dy}{dx} = -1(\sin(x) + \cos(x))^{-2} \cdot (\cos(x) - \sin(x))$ $\frac{dy}{dx} = -\frac{\cos(x) - \sin(x)}{(\sin(x) + \cos(x))^2} = \frac{\sin(x) - \cos(x)}{(\sin(x) + \cos(x))^2}$. Distractor D has a sign error in the numerator. The other distractors arise from incorrect differentiation methods, such as differentiating the numerator and denominator separately.
6. The point $(0, \frac{\pi}{4})$ lies on the curve defined by the equation $\tan(y) = x + \sec(x)$. What is the slope of the tangent line to the curve at this point?
   Answer: $\frac{1}{2}$ When you encounter an implicitly defined curve like this one, finding the slope requires implicit differentiation. The key insight is that you're looking for $\frac{dy}{dx}$ when the relationship between $x$ and $y$ isn't explicitly solved for one variable. Starting with $\tan(y) = x + \sec(x)$, differentiate both sides with respect to $x$. The left side gives $\sec^2(y) \cdot \frac{dy}{dx}$ using the chain rule. The right side becomes $1 + \sec(x)\tan(x)$. This yields: $\sec^2(y) \cdot \frac{dy}{dx} = 1 + \sec(x)\tan(x)$ Solving for the slope: $\frac{dy}{dx} = \frac{1 + \sec(x)\tan(x)}{\sec^2(y)}$ Now substitute the given point $(0, \frac{\pi}{4})$. At $x = 0$: $\sec(0) = 1$ and $\tan(0) = 0$, so the numerator becomes $1 + 1 \cdot 0 = 1$. At $y = \frac{\pi}{4}$: $\sec(\frac{\pi}{4}) = \sqrt{2}$, so $\sec^2(\frac{\pi}{4}) = 2$. Therefore: $\frac{dy}{dx} = \frac{1}{2}$ Looking at the wrong answers: (A) $2$ likely comes from flipping the fraction or miscomputing $\sec^2(\frac{\pi}{4})$. (B) $1$ results from forgetting to divide by $\sec^2(y)$ in the final step. (C) $\sqrt{2}$ appears if you use $\sec(\frac{\pi}{4})$ instead of its square in the denominator. Strategy tip: In implicit differentiation problems, always substitute the point values last, and remember that $\sec(\frac{\pi}{4}) = \sqrt{2}$, so $\sec^2(\frac{\pi}{4}) = 2$ — this relationship appears frequently on calculus exams.
7. For which value of $x$ in the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$ does the function $f(x) = x - \tan(x)$ have a horizontal tangent line?
   Answer: $x = 0$ When you see a question about horizontal tangent lines, you're looking for points where the derivative equals zero. A horizontal tangent means the slope is zero at that point. To find where $f(x) = x - \tan(x)$ has a horizontal tangent, you need to find where $f'(x) = 0$. Taking the derivative: $f'(x) = 1 - \sec^2(x)$. Setting this equal to zero: $1 - \sec^2(x) = 0$, which gives us $\sec^2(x) = 1$. Since $\sec(x) = \frac{1}{\cos(x)}$, this means $\frac{1}{\cos^2(x)} = 1$, so $\cos^2(x) = 1$. This occurs when $\cos(x) = \pm 1$. In the interval $(-\frac{\pi}{2}, \frac{\pi}{2})$, cosine equals 1 only at $x = 0$. Let's check each option: For choice A, when $x = 1$, $f'(1) = 1 - \sec^2(1) \neq 0$ since $\sec^2(1) > 1$. For choice B, when $x = \frac{\pi}{4}$, $f'(\frac{\pi}{4}) = 1 - \sec^2(\frac{\pi}{4}) = 1 - 2 = -1 \neq 0$. Choice D is incorrect because we found that such a value does exist. Choice C is correct: at $x = 0$, $f'(0) = 1 - \sec^2(0) = 1 - 1 = 0$. Remember: horizontal tangent lines always occur where the derivative equals zero. When working with trigonometric functions, pay special attention to the domain restrictions and use fundamental trigonometric identities to solve derivative equations systematically.
8. If $y = \tan(x^3 + 1)$, then $\frac{dy}{dx} =$
   Answer: $3x^2 \sec^2(x^3+1)$ This requires the chain rule. Let the outer function be $f(u) = \tan(u)$ and the inner function be $u(x) = x^3 + 1$. The derivative is $f'(u(x)) \cdot u'(x)$. $f'(u) = \sec^2(u) \implies f'(u(x)) = \sec^2(x^3+1)$. $u'(x) = 3x^2$. Combining these gives $\frac{dy}{dx} = \sec^2(x^3+1) \cdot 3x^2 = 3x^2 \sec^2(x^3+1)$. Distractor B misapplies the chain rule. Distractor C omits the derivative of the inner function. Distractor D uses the derivative rule for secant instead of tangent.
9. Let $f(x) = \cot(x)$. What is the value of the second derivative, $f''(\pi/4)$?
   Answer: 4 First, find the first derivative of $f(x) = \cot(x)$: $f'(x) = -\csc^2(x) = -(\csc(x))^2$. Now, find the second derivative using the chain rule: $f''(x) = -2(\csc(x)) \cdot \frac{d}{dx}(\csc(x))$ $f''(x) = -2\csc(x) \cdot (-\csc(x)\cot(x)) = 2\csc^2(x)\cot(x)$. Evaluate $f''(\pi/4)$: $f''(\pi/4) = 2\csc^2(\pi/4)\cot(\pi/4)$. Since $\csc(\pi/4) = \sqrt{2}$ and $\cot(\pi/4) = 1$: $f''(\pi/4) = 2(\sqrt{2})^2(1) = 2(2)(1) = 4$. Distractor A has a sign error. Distractor D results from forgetting to square the cosecant term.
10. If $y = \sec(x)$, which of the following expressions is equivalent to $\frac{d^2y}{dx^2}$?
    Answer: $2y^3 - y$ First, find the first derivative: $\frac{dy}{dx} = \sec(x)\tan(x)$. Next, use the product rule to find the second derivative: $\frac{d^2y}{dx^2} = (\sec(x)\tan(x))\tan(x) + \sec(x)(\sec^2(x)) = \sec(x)\tan^2(x) + \sec^3(x)$. Now, we must express this in terms of $y = \sec(x)$. We use the identity $\tan^2(x) = \sec^2(x) - 1 = y^2 - 1$. Substitute $\sec(x) = y$ and $\tan^2(x) = y^2 - 1$ into the expression for the second derivative: $\frac{d^2y}{dx^2} = y(y^2 - 1) + y^3 = y^3 - y + y^3 = 2y^3 - y$. Distractor C results from using the incorrect identity $\tan^2(x) = \sec^2(x) + 1$. Other distractors come from algebraic errors.
11. What is the equation of the line tangent to the graph of $g(x) = \cot(\pi x)$ at $x = 1/4$?
    Answer: $y - 1 = -2\pi(x - 1/4)$ First, find the point of tangency. At $x = 1/4$, $g(1/4) = \cot(\pi/4) = 1$. The point is $(1/4, 1)$. Next, find the derivative to determine the slope. Using the chain rule, $g'(x) = -\csc^2(\pi x) \cdot \pi$. Evaluate the derivative at $x = 1/4$ to find the slope $m$: $m = g'(1/4) = -\pi \csc^2(\pi/4) = -\pi (\sqrt{2})^2 = -2\pi$. Using the point-slope form $y - y_1 = m(x - x_1)$, the equation of the tangent line is $y - 1 = -2\pi(x - 1/4)$. Distractor A results from forgetting the chain rule factor of $\pi$. Distractor B has a sign error in the derivative of cotangent. Distractor D uses an incorrect value for cot(pi/4).
12. Let $f(x) = \sec(x)\cot(x)$. Which of the following is equivalent to $f'(x)$ for all $x$ in the domain of $f$?
    Answer: $-\csc(x)\cot(x)$ The most efficient method is to simplify $f(x)$ before differentiating. $f(x) = \sec(x)\cot(x) = \frac{1}{\cos(x)} \cdot \frac{\cos(x)}{\sin(x)} = \frac{1}{\sin(x)} = \csc(x)$. Now, we can find the derivative of the simplified function: $f'(x) = \frac{d}{dx}(\csc(x)) = -\csc(x)\cot(x)$. Alternatively, using the product rule on the original function gives $f'(x) = (\sec(x)\tan(x))\cot(x) + \sec(x)(-\csc^2(x)) = \sec(x) - \sec(x)\csc^2(x)$, which simplifies to the same result. Distractor C represents an unsimplified application of the product rule. Distractor A results from an incorrect differentiation rule.
13. At $x = \pi/4$, which of the following functions has the greatest instantaneous rate of change?
    Answer: $f(x) = \tan(x)$ We need to find the derivative of each function and evaluate it at $x = \pi/4$. For A: $f'(x) = \sec^2(x)$. $f'(\pi/4) = \sec^2(\pi/4) = (\sqrt{2})^2 = 2$. For B: $g'(x) = -\csc^2(x)$. $g'(\pi/4) = -\csc^2(\pi/4) = -(\sqrt{2})^2 = -2$. For C: $h'(x) = \sec(x)\tan(x)$. $h'(\pi/4) = \sec(\pi/4)\tan(\pi/4) = (\sqrt{2})(1) = \sqrt{2} \approx 1.414$. For D: $k'(x) = -\csc(x)\cot(x)$. $k'(\pi/4) = -\csc(\pi/4)\cot(\pi/4) = -(\sqrt{2})(1) = -\sqrt{2} \approx -1.414$. Comparing the values $2, -2, \sqrt{2}, -\sqrt{2}$, the greatest value is 2.
14. The velocity of a particle moving along a line is given by $v(t) = 10\tan(\frac{\pi t}{8})$ for $0 \le t < 4$. What is the acceleration of the particle at $t=2$?
    Answer: $5\pi/2$ Acceleration $a(t)$ is the derivative of velocity $v(t)$. We must use the chain rule to find $v'(t)$. $a(t) = v'(t) = 10 \cdot \sec^2(\frac{\pi t}{8}) \cdot \frac{d}{dt}(\frac{\pi t}{8})$ $a(t) = 10 \sec^2(\frac{\pi t}{8}) \cdot \frac{\pi}{8} = \frac{10\pi}{8}\sec^2(\frac{\pi t}{8}) = \frac{5\pi}{4}\sec^2(\frac{\pi t}{8})$. Now, evaluate the acceleration at $t=2$: $a(2) = \frac{5\pi}{4}\sec^2(\frac{2\pi}{8}) = \frac{5\pi}{4}\sec^2(\frac{\pi}{4})$. Since $\sec(\pi/4) = \sqrt{2}$, $\sec^2(\pi/4) = 2$. $a(2) = \frac{5\pi}{4} \cdot 2 = \frac{10\pi}{4} = \frac{5\pi}{2}$. Distractor A results from forgetting the chain rule factor of $\pi/8$. Distractor C is an intermediate calculation before substituting the value of $\sec^2(\pi/4)$.
15. Let $f(x) = x^2 \sec(x)$. What is the value of $f'(\pi)$?
    Answer: $-2\pi$ To find $f'(x)$, we use the product rule: $(uv)' = u'v + uv'$. Let $u = x^2$ and $v = \sec(x)$. Then $u' = 2x$ and $v' = \sec(x)\tan(x)$. So, $f'(x) = (2x)(\sec(x)) + (x^2)(\sec(x)\tan(x))$. Now, we evaluate $f'(x)$ at $x = \pi$. $f'(\pi) = 2\pi \sec(\pi) + (\pi)^2 \sec(\pi)\tan(\pi)$. We know that $\sec(\pi) = -1$ and $\tan(\pi) = 0$. Substituting these values: $f'(\pi) = 2\pi(-1) + \pi^2(-1)(0) = -2\pi + 0 = -2\pi$.
16. Let $f(x) = \csc(x)$. What is the value of the second derivative, $f''(\frac{\pi}{2})$?
    Answer: $1$ When you encounter a problem asking for the second derivative of a trigonometric function, you need to systematically find the first derivative, then differentiate again. Starting with $f(x) = \csc(x)$, recall that $\csc(x) = \frac{1}{\sin(x)}$. The derivative of cosecant is $f'(x) = -\csc(x)\cot(x)$. To find the second derivative, you'll need to use the product rule on $f'(x) = -\csc(x)\cot(x)$: $f''(x) = -[\csc'(x)\cot(x) + \csc(x)\cot'(x)]$ Since $\csc'(x) = -\csc(x)\cot(x)$ and $\cot'(x) = -\csc^2(x)$: $f''(x) = -[(-\csc(x)\cot(x))\cot(x) + \csc(x)(-\csc^2(x))]$ $f''(x) = \csc(x)\cot^2(x) + \csc^3(x) = \csc(x)[\cot^2(x) + \csc^2(x)]$ Now evaluate at $x = \frac{\pi}{2}$. At this point, $\sin(\frac{\pi}{2}) = 1$, so $\csc(\frac{\pi}{2}) = 1$, and $\cos(\frac{\pi}{2}) = 0$, so $\cot(\frac{\pi}{2}) = 0$. Therefore: $f''(\frac{\pi}{2}) = 1[0^2 + 1^2] = 1$ Answer choice D is correct. Choice A (2) might result from computational errors in the product rule. Choice B (-1) could come from sign errors during differentiation. Choice C (0) might arise from incorrectly thinking that since $\cot(\frac{\pi}{2}) = 0$, the entire expression equals zero. Remember: when differentiating products of trigonometric functions, work methodically through each step and be especially careful with signs and trigonometric values at special angles.
17. What is the derivative with respect to $x$ of the function $f(x) = \csc(e^{2x})$?
    Answer: $-2e^{2x} \csc(e^{2x})\cot(e^{2x})$ This requires a double application of the chain rule. The outermost function is $\csc(u)$, the middle is $e^v$, and the innermost is $2x$. Let $u = e^{2x}$. Then $f(x) = \csc(u)$. $f'(x) = \frac{df}{du} \cdot \frac{du}{dx}$. $\frac{df}{du} = -\csc(u)\cot(u) = -\csc(e^{2x})\cot(e^{2x})$. To find $\frac{du}{dx}$, we use the chain rule again for $e^{2x}$: $\frac{du}{dx} = e^{2x} \cdot \frac{d}{dx}(2x) = 2e^{2x}$. Combining the parts: $f'(x) = -\csc(e^{2x})\cot(e^{2x}) \cdot 2e^{2x} = -2e^{2x} \csc(e^{2x})\cot(e^{2x})$. Distractor C has a sign error. Distractors B and D result from incomplete application of the chain rule.
18. Let $g(x) = \csc(f(x))$. If $f(2) = \pi/6$ and $f'(2) = 4$, what is the value of $g'(2)$?
    Answer: $-8\sqrt{3}$ We use the chain rule to find $g'(x)$: $g'(x) = -\csc(f(x))\cot(f(x)) \cdot f'(x)$. Now, we substitute the given values at $x=2$: $g'(2) = -\csc(f(2))\cot(f(2)) \cdot f'(2)$. Given $f(2) = \pi/6$ and $f'(2) = 4$: $g'(2) = -\csc(\pi/6)\cot(\pi/6) \cdot 4$. We know $\csc(\pi/6) = 2$ and $\cot(\pi/6) = \sqrt{3}$. $g'(2) = -(2)(\sqrt{3}) \cdot 4 = -8\sqrt{3}$. Distractor B has a sign error. Distractor A results from forgetting the $f'(2)$ factor. Distractor C results from using the trigonometric values for $\pi/3$ instead of $\pi/6$.
19. The equation of the tangent line to the graph of $f(x) = \cot(ax)$ at $x = \frac{\pi}{4a}$ is given by $y - 1 = -4(x - \frac{\pi}{4a})$. What is the value of the constant $a$?
    Answer: 2 From the tangent line equation, the slope is $m = -4$. The slope is also given by the derivative $f'(x)$ at the point of tangency. First, find the derivative of $f(x) = \cot(ax)$ using the chain rule: $f'(x) = -\csc^2(ax) \cdot a = -a\csc^2(ax)$. Evaluate the derivative at $x = \frac{\pi}{4a}$: $f'(\frac{\pi}{4a}) = -a\csc^2(a \cdot \frac{\pi}{4a}) = -a\csc^2(\frac{\pi}{4})$. Since $\csc(\pi/4) = \sqrt{2}$, we have $\csc^2(\pi/4) = 2$. So, $f'(\frac{\pi}{4a}) = -a(2) = -2a$. Set this equal to the given slope: $-2a = -4$. Solving for $a$ gives $a = 2$. Distractor C results from forgetting to square $\csc(\pi/4)$. Distractor D is a sign error.
20. Let $f(x) = \sec^3(\frac{x}{3})$. What is the value of $f'(\pi)$?
    Answer: $8\sqrt{3}$ When you encounter a derivative problem involving composite functions like $f(x) = \sec^3(\frac{x}{3})$, you need to apply the chain rule systematically, often multiple times. To find $f'(x)$, start by recognizing this as a composition: an outer function (cubing) applied to $\sec(\frac{x}{3})$, which itself involves the inner function $\frac{x}{3}$. Using the chain rule: $f'(x) = 3\sec^2(\frac{x}{3}) \cdot \frac{d}{dx}[\sec(\frac{x}{3})]$ The derivative of $\sec(u)$ is $\sec(u)\tan(u)$, so: $\frac{d}{dx}[\sec(\frac{x}{3})] = \sec(\frac{x}{3})\tan(\frac{x}{3}) \cdot \frac{1}{3}$ Therefore: $f'(x) = 3\sec^2(\frac{x}{3}) \cdot \sec(\frac{x}{3})\tan(\frac{x}{3}) \cdot \frac{1}{3} = \sec^3(\frac{x}{3})\tan(\frac{x}{3})$ Now evaluate at $x = \pi$: $f'(\pi) = \sec^3(\frac{\pi}{3})\tan(\frac{\pi}{3})$ Since $\sec(\frac{\pi}{3}) = 2$ and $\tan(\frac{\pi}{3}) = \sqrt{3}$: $f'(\pi) = 2^3 \cdot \sqrt{3} = 8\sqrt{3}$ This confirms answer B is correct. Answer A ($24\sqrt{3}$) likely comes from incorrectly keeping the factor of 3 instead of $\frac{1}{3}$. Answer C ($4$) misses the $\sqrt{3}$ factor entirely. Answer D ($16$) probably results from calculation errors with the trigonometric values. Study tip: For composite trigonometric functions, write out each step of the chain rule carefully and double-check your trigonometric values at special angles like $\frac{\pi}{3}$.

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