Calculus 1 Quiz: Derivatives Of Tan Cot Sec Csc

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Derivatives Of Tan Cot Sec CscQuestion 1 of 20

What is an equation of the line tangent to the graph of $y = 2\csc(x)$ at the point where $x = \frac{5\pi}{6}$ ?

y + 4 = -4\sqrt{3}(x - \frac{5\pi}{6})

y - 4 = -4\sqrt{3}(x - \frac{5\pi}{6})

y - 4 = 4\sqrt{3}(x - \frac{5\pi}{6})

y - 4 = -\sqrt{3}(x - \frac{5\pi}{6})

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Calculus 1 Quiz: Derivatives Of Tan Cot Sec Csc

Practice Derivatives Of Tan Cot Sec Csc in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Derivatives Of Tan Cot Sec Csc, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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Question 1

What is an equation of the line tangent to the graph of $y = 2\csc(x)$ at the point where $x = \frac{5\pi}{6}$ ?

$y + 4 = -4\sqrt{3}(x - \frac{5\pi}{6})$
$y - 4 = -4\sqrt{3}(x - \frac{5\pi}{6})$
$y - 4 = 4\sqrt{3}(x - \frac{5\pi}{6})$ (correct answer)
$y - 4 = -\sqrt{3}(x - \frac{5\pi}{6})$

Explanation: 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)

-\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.

Question 2

What is the slope of the normal line to the curve $y = 3\cot(x/2)$ at $x = \pi$ ?

$-3/2$
$2/3$ (correct answer)
$-2/3$
$1/3$

Explanation: First, find the derivative to get the slope of the tangent line,

m_{\text{tan}}

. Using the chain rule:

\frac{dy}{dx} = 3 \cdot [-\csc^2(x/2)] \cdot \frac{1}{2} = -\frac{3}{2}\csc^2(x/2)

. Evaluate the derivative at

x = \pi

m_{\text{tan}} = -\frac{3}{2}\csc^2(\pi/2) = -\frac{3}{2}(1)^2 = -\frac{3}{2}

. The slope of the normal line,

m_{\text{norm}}

, is the negative reciprocal of the tangent slope:

m_{\text{norm}} = -\frac{1}{m_{\text{tan}}} = -\frac{1}{-3/2} = \frac{2}{3}

. Distractor A is the slope of the tangent line. Distractor C is the reciprocal, but not the negative reciprocal. Distractor D results from forgetting the chain rule factor of 1/2.

Question 3

What is the derivative with respect to $x$ of the function $f(x) = \csc(e^{2x})$ ?

$-2e^{2x} \csc(e^{2x})\cot(e^{2x})$ (correct answer)
$-e^{2x} \csc(e^{2x})\cot(e^{2x})$
$2e^{2x} \csc(e^{2x})\cot(e^{2x})$
$-\csc(e^{2x})\cot(e^{2x})$

Explanation: 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.

Question 4

Let $h(x) = \tan(x) \sec(2x)$ . What is the value of $h'(0)$ ?

0
1 (correct answer)
2
3

Explanation: 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.

Question 5

Let $f(x) = \tan(x)$ and $g(x) = \sec(x)$ . Which of the following statements is true for all $x$ in the interval $(0, \pi/2)$ ?

$f'(x) = g(x)$
$g'(x) = f(x)$
$f'(x) = [g(x)]^2$ (correct answer)
$g'(x) = [f(x)]^2$

Explanation: We need to find the derivatives of

f(x)

and

g(x)

and express them in terms of

f(x)

and

g(x)

. The derivative of

f(x) = \tan(x)

f'(x) = \sec^2(x)

. Since

g(x) = \sec(x)

, we have

\sec^2(x) = (\sec(x))^2 = [g(x)]^2

. Thus,

f'(x) = [g(x)]^2

, making statement C true. The derivative of

g(x) = \sec(x)

g'(x) = \sec(x)\tan(x)

. In terms of f and g, this is

g'(x) = g(x)f(x)

. This shows that statements B and D are false. Statement A is also false, as

\sec^2(x) \neq \sec(x)

in general.

Question 6

At $x = \pi/4$ , which of the following functions has the greatest instantaneous rate of change?

$f(x) = \tan(x)$ (correct answer)
$g(x) = \cot(x)$
$h(x) = \sec(x)$
$k(x) = \csc(x)$

Explanation: 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.

Question 7

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?

$2$
$1$
$\sqrt{2}$
$\frac{1}{2}$ (correct answer)

Explanation: 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.

Question 8

For how many values of $x$ in the interval $(0, 2\pi)$ is the slope of the tangent line to the graph of $f(x) = \cot(x)$ equal to $-1$ ?

Zero
Two (correct answer)
One
Four

Explanation: This question tests your ability to find where a derivative equals a specific value, combining trigonometric differentiation with equation solving. To find where the slope equals -1, you need to solve

f'(x) = -1

where

f(x) = \cot(x)

. The derivative of cotangent is

f'(x) = -\csc^2(x)

, so you're solving:

-\csc^2(x) = -1

Dividing by -1 gives

\csc^2(x) = 1

, which means

\frac{1}{\sin^2(x)} = 1

. This simplifies to

\sin^2(x) = 1

, so

\sin(x) = \pm 1

. In the interval

(0, 2\pi)

, sine equals 1 at

x = \frac{\pi}{2}

and equals -1 at

x = \frac{3\pi}{2}

. However, you must check that cotangent is actually defined at these points. Since

\cot(x) = \frac{\cos(x)}{\sin(x)}

, cotangent is undefined wherever sine equals zero, but it's perfectly defined where sine equals ±1. At both

x = \frac{\pi}{2}

and

x = \frac{3\pi}{2}

, we have

\csc^2(x) = 1

, so

f'(x) = -1

. Choice A (Zero) is wrong because we found valid solutions. Choice C (One) misses one of the two solutions. Choice D (Four) likely comes from incorrectly thinking about the period of cotangent or confusing this with a different trigonometric equation. When solving trigonometric equations involving derivatives, always verify that the original function is defined at your solution points, and carefully consider the given interval to count solutions accurately.

Question 9

A searchlight on the ground shines on a tall building 30 meters away. The height of the light spot on the building, $h$ , is given by $h = 30\tan(\theta)$ , where $\theta$ is the angle of elevation of the searchlight. What is the rate of change of the height with respect to the angle, $\frac{dh}{d\theta}$ , when the spot of light is $30\sqrt{3}$ meters high?

$60$ m/rad
$120$ m/rad (correct answer)
$60\sqrt{3}$ m/rad
$7.5$ m/rad

Explanation: This is a related rates problem that combines trigonometry with differentiation. When you see a geometric setup involving angles and changing quantities, you'll typically need to differentiate a trigonometric relationship. Given

h = 30\tan(\theta)

, you need to find

\frac{dh}{d\theta}

when the light spot is

30\sqrt{3}

meters high. Start by differentiating:

\frac{dh}{d\theta} = 30\sec^2(\theta)

Next, determine the angle

\theta

when

h = 30\sqrt{3}

30\sqrt{3} = 30\tan(\theta)

\tan(\theta) = \sqrt{3}

This means

\theta = 60°

\frac{\pi}{3}

radians, where

\sec(\theta) = \frac{1}{\cos(\theta)} = \frac{1}{\cos(60°)} = \frac{1}{1/2} = 2

Therefore:

\frac{dh}{d\theta} = 30\sec^2(60°) = 30(2)^2 = 30(4) = 120

m/rad Choice A (60 m/rad) likely comes from forgetting to square the secant:

30\sec(60°) = 30(2) = 60

. Choice C (

60\sqrt{3}

m/rad) probably results from incorrectly using

\tan(60°) = \sqrt{3}

in the derivative calculation. Choice D (7.5 m/rad) might stem from a computational error or misapplying the chain rule. Strategy tip: In related rates problems involving trigonometry, always remember that the derivative of

\tan(x)

\sec^2(x)

, not just

\sec(x)

. Write out your trigonometric derivatives carefully and double-check which angle corresponds to your given conditions.

Question 10

Let $f(x) = e^x \sec(x)$ . What is the value of $f'(0)$ ?

0
1 (correct answer)
$e$
The derivative does not exist.

Explanation: 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)

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

Question 11

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 $a$ ?

4
2 (correct answer)
$2\sqrt{2}$
-2

Explanation: 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.

Question 12

If $y = \sec(x)$ , which of the following expressions is equivalent to $\frac{d^2y}{dx^2}$ ?

$2y^3 - y$ (correct answer)
$y^3 - y$
$2y^3 + y$
$y^2 - 1 + y^3$

Explanation: 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.

Question 13

Let $f(x) = x^2 \sec(x)$ . What is the value of $f'(\pi)$ ?

$-2\pi$ (correct answer)
$2\pi$
$0$
$-\pi^2$

Explanation: 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)

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

Question 14

If $h(x) = \tan(\cos(x))$ , what is $h'(x)$ ?

$-\sin(x)\sec^2(\cos(x))$ (correct answer)
$\sin(x)\sec^2(\cos(x))$
$-\sin(x)\sec^2(x)$
$\sec(\cos(x))\tan(\cos(x))$

Explanation: This requires the chain rule. The outer function is

\tan(u)

and the inner function is

u = \cos(x)

. The derivative of the outer function is

\sec^2(u)

. The derivative of the inner function is

-\sin(x)

. According to the chain rule,

h'(x) = \frac{d}{du}[\tan(u)] \cdot \frac{du}{dx} = \sec^2(u) \cdot (-\sin(x))

. Substituting

u = \cos(x)

back, we get

h'(x) = \sec^2(\cos(x)) \cdot (-\sin(x)) = -\sin(x)\sec^2(\cos(x))

Question 15

Let $h(x) = f(\tan(x) + 1)$ for some differentiable function $f$ . If $f'(2) = 5$ , what is the value of $h'(\frac{\pi}{4})$ ?

$5\sqrt{2}$
$5$
$10$ (correct answer)
$2\sqrt{5}$

Explanation: This question tests the chain rule, which you'll need whenever you're differentiating a composite function - a function inside another function. To find

h'(x)

where

h(x) = f(\tan(x) + 1)

, you need to apply the chain rule: the derivative of the outer function times the derivative of the inner function. Let's identify the parts: the outer function is

f(u)

where

u = \tan(x) + 1

, and the inner function is

u = \tan(x) + 1

. Using the chain rule:

h'(x) = f'(\tan(x) + 1) \cdot \frac{d}{dx}[\tan(x) + 1]

The derivative of

\tan(x) + 1

\sec^2(x)

, so:

h'(x) = f'(\tan(x) + 1) \cdot \sec^2(x)

Now evaluate at

x = \frac{\pi}{4}

$\tan(\frac{\pi}{4}) = 1$ , so $\tan(\frac{\pi}{4}) + 1 = 2$
$\sec(\frac{\pi}{4}) = \frac{1}{\cos(\frac{\pi}{4})} = \frac{1}{\frac{\sqrt{2}}{2}} = \sqrt{2}$
Therefore $\sec^2(\frac{\pi}{4}) = 2$

h'(\frac{\pi}{4}) = f'(2) \cdot 2 = 5 \cdot 2 = 10

Choice A)

5\sqrt{2}

likely comes from forgetting to square the secant value. Choice B)

5

results from forgetting the chain rule entirely - just using

f'(2)

. Choice D)

2\sqrt{5}

appears to mix up the arithmetic incorrectly. Remember: when you see composite functions, immediately think chain rule. Always multiply by the derivative of the inner function!

Question 16

What is the slope of the line tangent to the graph of $g(x) = \frac{\cot(x)}{1+x}$ at $x = \frac{\pi}{2}$ ?

$-\frac{1}{1 + \pi/2}$ (correct answer)
$\frac{1}{1 + \pi/2}$
$-(1 + \frac{\pi}{2})$
$0$

Explanation: To find the slope of the tangent line, we need to compute the derivative

g'(x)

using the quotient rule:

(\frac{u}{v})' = \frac{u'v - uv'}{v^2}

. Let

u = \cot(x)

and

v = 1+x

. Then

u' = -\csc^2(x)

and

v' = 1

g'(x) = \frac{(-\csc^2(x))(1+x) - (\cot(x))(1)}{(1+x)^2}

. Now, evaluate at

x = \frac{\pi}{2}

. We have

\cot(\frac{\pi}{2}) = 0

and

\csc(\frac{\pi}{2}) = 1

g'(\frac{\pi}{2}) = \frac{(-\csc^2(\frac{\pi}{2}))(1+\frac{\pi}{2}) - (\cot(\frac{\pi}{2}))(1)}{(1+\frac{\pi}{2})^2} = \frac{(-(1)^2)(1+\frac{\pi}{2}) - 0}{(1+\frac{\pi}{2})^2} = \frac{-(1+\frac{\pi}{2})}{(1+\frac{\pi}{2})^2} = -\frac{1}{1+\frac{\pi}{2}}

Question 17

Let $f(x) = \sec^3(\frac{x}{3})$ . What is the value of $f'(\pi)$ ?

$24\sqrt{3}$
$8\sqrt{3}$ (correct answer)
$4$
$16$

Explanation: 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)

\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}

Question 18

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?

$x = 1$
$x = \frac{\pi}{4}$
$x = 0$ (correct answer)
No such value exists.

Explanation: 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.

Question 19

Let $f(x) = \csc(x)$ . What is the value of the second derivative, $f''(\frac{\pi}{2})$ ?

$2$
$-1$
$0$
$1$ (correct answer)

Explanation: 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.

Question 20

If $f(x) = \frac{\cot(x)}{x}$ , find $f'(\pi/2)$ .

$-\frac{2}{\pi}$ (correct answer)
$\frac{2}{\pi}$
$-\frac{4}{\pi^2}$
0

Explanation: We use the quotient rule:

\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}

. Here,

u = \cot(x)

and

v = x

. So

u' = -\csc^2(x)

and

v' = 1

f'(x) = \frac{-\csc^2(x) \cdot x - \cot(x) \cdot 1}{x^2}

. Now, evaluate at

x = \pi/2

f'(\pi/2) = \frac{-\csc^2(\pi/2) \cdot (\pi/2) - \cot(\pi/2)}{(\pi/2)^2}

. Since

\csc(\pi/2) = 1

and

\cot(\pi/2) = 0

, we have:

f'(\pi/2) = \frac{-(1)^2 \cdot (\pi/2) - 0}{\pi^2/4} = \frac{-\pi/2}{\pi^2/4} = -\frac{\pi}{2} \cdot \frac{4}{\pi^2} = -\frac{2}{\pi}

. Distractor B is a sign error. Distractor C results from an error in evaluating

\csc(\pi/2)

. Distractor D results from mis-evaluating the numerator.