Grade 9 Exam > Grade 9 Notes > Calculus AB > Chapter Notes: Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions
Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions Chapter Notes | Calculus AB - Grade 9 PDF Download
Welcome to your guide for conquering advanced trigonometric derivatives in AP Calculus! If you’ve already nailed the derivatives of sin(x) and cos(x), you’re ready to tackle the rest. Memorizing these rules will make your calculus journey smoother and more intuitive.
Overview of Trigonometric Derivatives
Below is a concise table summarizing the derivatives of tangent, cotangent, secant, and cosecant functions. Note that these derivatives apply only when angles are in radians, not degrees.
Tip: Familiarize yourself with trigonometric identities like tan(x) = sin(x)/cos(x) and cot(x) = 1/tan(x) to simplify expressions before differentiating.
Derivative of tan(x)
The derivative of tan(x) is sec²(x). Let’s explore an example:
Consider the function: f(x) = 3tan(x) + 2x²
Differentiate each term separately:
- For 3tan(x), the derivative is 3sec²(x).
- For 2x², the derivative is 4x.
Thus, the derivative is: f'(x) = 3sec²(x) + 4x.
Derivative of cot(x)
The derivative of cot(x) is -csc²(x). Here’s an example:
Given: f(x) = 5cot(x) + x
We again have to differentiate the two terms separately! The derivative of cot x is −csc2x, so the derivative of the first term is −5csc2x. The derivative of x is 1. Therefore, f′(x)=−5csc2x + 1 or f′(x) = 1 − 5csc2x.
Derivative of sec(x)
The derivative of sec(x) is sec(x)tan(x). For example:
Take: f(x) = 2sec(x) + 3x³
Knowing the above trig derivative rule, the derivative of the first term is 2secx tanx. The derivative of 3x3 is 9x2. Thus, f′(x) = 2secx tanx + 9x2.
Derivative of csc(x)
The derivative of csc(x) is -csc(x)cot(x). Here’s an example:
Consider: f(x) = 4csc(x) + 7x²
Differentiate each part:
- The derivative of 4csc(x) is -4csc(x)cot(x).
- The derivative of 7x² is 14x.
Therefore: f'(x) = -4csc(x)cot(x) + 14x.
Practice Problems
Test your understanding with these practice questions. Apply the chain rule, sum rule, and quotient rule where necessary.
- f(x) = 2tan(x) + sec(x)
- f(x) = cot(x)/csc(x)
- g(x) = tan²(6x)
- h(x) = 5cot(x)
Solutions
- f'(x) = 2sec²(x) + sec(x)tan(x)
- f'(x) = -csc²(x)
- g'(x) = 2tan(6x)(1/cos²(6x))
- h'(x) = -5csc²(x)
These problems integrate multiple derivative rules. If you need a refresher, review the power rule, sum/difference rules, product rule, quotient rule, and derivatives of basic functions like sin(x), cos(x), ex, and ln(x).
Question for Chapter Notes: Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant FunctionsTry yourself: What is the derivative of tan(x)?View Solution
Key Terms to Understand
- Chain Rule: A method for differentiating composite functions, multiplying the derivative of the outer function by the derivative of the inner function.
- Cotangent: A trigonometric ratio defined as the adjacent side over the opposite side in a right triangle.
- (cot(x))': The derivative of the cotangent function, indicating its rate of change.
- -csc²(x): The derivative of the cosecant function squared, with a negative sign.
- -csc(x)cot(x): The product of cosecant and cotangent functions, negated, used in derivative calculations.
- Derivatives: Measures of how a function changes as its input varies.
- sec²(x): The derivative of the secant function, representing its rate of change.
- Secant: A trigonometric ratio defined as the hypotenuse over the adjacent side in a right triangle.
- sec(x)': The derivative of the secant function, showing its rate of change.
- Tangent: A line that touches a curve at one point, sharing the same slope as the curve at that point.
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FAQs on Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions Chapter Notes - Calculus AB - Grade 9
| 1. What are the basic derivatives of trigonometric functions? |  |
Ans. The basic derivatives of trigonometric functions are as follows:
- The derivative of sin(x) is cos(x).
- The derivative of cos(x) is -sin(x).
- The derivative of tan(x) is sec^2(x).
- The derivative of cot(x) is -csc^2(x).
- The derivative of sec(x) is sec(x)tan(x).
- The derivative of csc(x) is -csc(x)cot(x).
| 2. How do you find the derivative of the tangent function? | |
Ans. To find the derivative of the tangent function, you use the formula:
The derivative of tan(x) is sec^2(x). This can be derived from the quotient rule since tan(x) is sin(x)/cos(x), or you can memorize this derivative as a key trigonometric derivative.
| 3. What is the derivative of cotangent, and how is it derived? | |
Ans. The derivative of cotangent is -csc^2(x). This can be derived by applying the quotient rule to cot(x) = cos(x)/sin(x), which results in the derivative being -csc^2(x) after simplification.
| 4. Can you explain how to differentiate secant and cosecant functions? | |
Ans. Yes! The derivative of sec(x) is sec(x)tan(x), which can be understood by applying the product rule and the chain rule. For cosecant, the derivative is -csc(x)cot(x), which is also derived using the quotient rule.
| 5. Why are trigonometric derivatives important in calculus? | |
Ans. Trigonometric derivatives are important in calculus because they help in solving problems involving rates of change, optimization, and modeling periodic phenomena. They are essential in higher-level mathematics and physics, where understanding the behavior of trigonometric functions is crucial.
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Oct 05, 2025
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