Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions - AP Calculus BC Study Guide

The Core Idea: Finding the Derivatives of Tangent, Cotangent, Secant, and/or Cosecant Functions

This topic expands our differentiation toolkit to include all six standard trigonometric functions. While the derivatives of sine and cosine are foundational, the derivatives of tangent, cotangent, secant, and cosecant are not arbitrary rules to be memorized in isolation. The core concept is that these four additional derivative rules can be logically derived using the definitions of the functions in terms of sine and cosine (e.g., $\tan (x)=\frac{\sin (x)}{\cos (x)}$) in conjunction with the quotient rule.

Understanding this connection is crucial. It reinforces the interconnectedness of calculus rules and provides a method to re-derive the formulas if they are forgotten. By mastering these derivatives, we can analyze the rates of change of a wider variety of periodic functions and solve more complex problems involving trigonometric models.

Key Formulas

The derivatives of the tangent, cotangent, secant, and cosecant functions are essential rules for differentiation.

• Derivative of Tangent:

  $$\frac{d}{dx}(\tan (x))={\sec }^2(x)$$

• Derivative of Cotangent:

  $$\frac{d}{dx}(\cot (x))=-{\csc }^2(x)$$

• Derivative of Secant:

  $$\frac{d}{dx}(\sec (x))=\sec (x)\tan (x)$$

• Derivative of Cosecant:

  $$\frac{d}{dx}(\csc (x))=-\csc (x)\cot (x)$$

Understanding the Derivations

A key piece of essential knowledge is that these formulas are not random; they are direct consequences of the derivatives of sine and cosine, applied through the quotient rule. Understanding how to derive them provides a deeper conceptual grasp and a fallback method for recalling the formulas.

Let's demonstrate the derivation for $\frac{d}{dx}(\tan (x))$.

1. Rewrite the function: Start by expressing $\tan (x)$ in terms of $\sin (x)$ and $\cos (x)$.

   $$\tan (x)=\frac{\sin (x)}{\cos (x)}$$

2. Apply the Quotient Rule: Use the quotient rule, $\frac{d}{dx}\left(\frac{f}{g}\right)=\frac{g\cdot f^{\prime }-f\cdot g^{\prime }}{g^2}$, where $f(x)=\sin (x)$ and $g(x)=\cos (x)$. We know $f^{\prime }(x)=\cos (x)$ and $g^{\prime }(x)=-\sin (x)$.

   $$\frac{d}{dx}(\tan (x))=\frac{(\cos (x))\cdot \frac{d}{dx}(\sin (x))-(\sin (x))\cdot \frac{d}{dx}(\cos (x))}{(\cos (x){)}^2}$$

3. Substitute the known derivatives:

   $$\frac{d}{dx}(\tan (x))=\frac{(\cos (x))(\cos (x))-(\sin (x))(-\sin (x))}{{\cos }^2(x)}$$

4. Simplify the numerator:

   $$\frac{d}{dx}(\tan (x))=\frac{{\cos }^2(x)+{\sin }^2(x)}{{\cos }^2(x)}$$

5. Apply the Pythagorean Identity: Use the identity ${\sin }^2(x)+{\cos }^2(x)=1$.

   $$\frac{d}{dx}(\tan (x))=\frac{1}{{\cos }^2(x)}$$

6. Use the reciprocal identity: Since $\sec (x)=\frac{1}{\cos (x)}$, we have $\frac{1}{{\cos }^2(x)}={\sec }^2(x)$.

   $$\frac{d}{dx}(\tan (x))={\sec }^2(x)$$

This same process can be used to derive the formulas for $\cot (x)$, $\sec (x)$, and $\csc (x)$.

Core Concepts & Rules

• The derivatives of all six trigonometric functions must be known. The four covered in this topic are:

  • $\frac{d}{dx}(\tan (x))={\sec }^2(x)$

  • $\frac{d}{dx}(\cot (x))=-{\csc }^2(x)$

  • $\frac{d}{dx}(\sec (x))=\sec (x)\tan (x)$

  • $\frac{d}{dx}(\csc (x))=-\csc (x)\cot (x)$

• Each of these derivative rules can be proven by first rewriting the function in terms of sine and cosine and then applying the quotient rule.

• Notice a pattern with the "co-" functions: the derivatives of $\cot (x)$ and $\csc (x)$ are negative.

Step-by-Step Example 1: Basic Application

Problem: Find the derivative of $f(x)=2\sec (x)-\cot (x)$.

Solution:

1. Apply the Difference Rule: The derivative of a difference is the difference of the derivatives.

   $$f^{\prime }(x)=\frac{d}{dx}(2\sec (x))-\frac{d}{dx}(\cot (x))$$

2. Apply the Constant Multiple Rule: The constant $2$ can be factored out of the first derivative.

   $$f^{\prime }(x)=2\cdot \frac{d}{dx}(\sec (x))-\frac{d}{dx}(\cot (x))$$

3. Apply the Specific Trigonometric Derivative Rules: Use the memorized formulas for the derivatives of secant and cotangent.

   • $\frac{d}{dx}(\sec (x))=\sec (x)\tan (x)$

   • $\frac{d}{dx}(\cot (x))=-{\csc }^2(x)$

   Substitute these into the expression:

   $$f^{\prime }(x)=2(\sec (x)\tan (x))-(-{\csc }^2(x))$$

4. Simplify the Expression: Be careful with the double negative.

   $$f^{\prime }(x)=2\sec (x)\tan (x)+{\csc }^2(x)$$

Step-by-Step Example 2: Exam-Style Application

Problem: Find the equation of the line normal to the graph of $y=x\csc (x)$ at $x=\frac{\pi }{6}$.

Solution:

1. Find the derivative, $y^{\prime }$: The function is a product of $x$ and $\csc (x)$, so we must use the product rule: $(fg{)}^{\prime }=f^{\prime }g+f{g}^{\prime }$.

   • Let $f(x)=x$ and $g(x)=\csc (x)$.

   • Then $f^{\prime }(x)=1$ and $g^{\prime }(x)=-\csc (x)\cot (x)$.

   $$y^{\prime }=\frac{d}{dx}(x\csc (x))=(1)(\csc (x))+(x)(-\csc (x)\cot (x))$$

   $$y^{\prime }=\csc (x)-x\csc (x)\cot (x)$$

2. Find the slope of the tangent line ($m_{tan}$): Evaluate the derivative $y^{\prime }$ at $x=\frac{\pi }{6}$.

   • Recall the values from the unit circle:

     • $\sin (\frac{\pi }{6})=\frac{1}{2}⟹\csc (\frac{\pi }{6})=2$

     • $\cos (\frac{\pi }{6})=\frac{\sqrt{3}}{2}$

     • $\tan (\frac{\pi }{6})=\frac{1}{\sqrt{3}}⟹\cot (\frac{\pi }{6})=\sqrt{3}$

   Substitute these values into $y^{\prime }$:

   $$m_{tan}=\csc \left(\frac{\pi }{6}\right)-\frac{\pi }{6}\csc \left(\frac{\pi }{6}\right)\cot \left(\frac{\pi }{6}\right)$$

   $$m_{tan}=2-\frac{\pi }{6}(2)(\sqrt{3})$$

   $$m_{tan}=2-\frac{\pi \sqrt{3}}{3}$$

3. Find the slope of the normal line ($m_{norm}$): The normal line is perpendicular to the tangent line. Its slope is the opposite reciprocal of $m_{tan}$.

   $$m_{norm}=-\frac{1}{m_{tan}}=-\frac{1}{2-\frac{\pi \sqrt{3}}{3}}$$

4. Find the point of tangency ($(x_1,y_1)$): Evaluate the original function $y=x\csc (x)$ at $x=\frac{\pi }{6}$.

   $$y_1=\frac{\pi }{6}\csc \left(\frac{\pi }{6}\right)=\frac{\pi }{6}(2)=\frac{\pi }{3}$$

   The point is $(\frac{\pi }{6},\frac{\pi }{3})$.

5. Write the equation of the normal line: Use the point-slope form $y-y_1=m(x-x_1)$.

   $$y-\frac{\pi }{3}=-\frac{1}{2-\frac{\pi \sqrt{3}}{3}}\left(x-\frac{\pi }{6}\right)$$

Using Your Calculator

The derivatives of tangent, cotangent, secant, and cosecant are found analytically (by hand). A calculator is not used to find the symbolic derivative itself, but it is an excellent tool for checking your work.

To verify the derivative found in Example 2, $y^{\prime }=\csc (x)-x\csc (x)\cot (x)$ for $y=x\csc (x)$:

1. Graph the numerical derivative: In your calculator's graphing screen (e.g., Y1), enter the numerical derivative of the original function. The syntax is typically nDeriv( or $d/dx($.

   • Y1 = nDeriv(X * 1/sin(X), X, X)

   (Note: You must enter $\csc (x)$ as $1/\sin (x)$)

2. Graph your analytical derivative: In a second slot (e.g., Y2), enter the derivative you calculated by hand.

   • Y2 = 1/sin(X) - X * (1/sin(X)) * (1/tan(X))

3. Compare: Graph both functions. If they produce the exact same curve, your analytical derivative is very likely correct. You can also compare their values in the table (TBLSET/TABLE) to confirm they match for all $x$-values.

AP Exam Quick Hit

Common Question Types

• Combining with Product/Quotient/Chain Rules: You will rarely be asked for just the derivative of $\tan (x)$. More commonly, you'll need to find the derivative of a more complex function.

  • Example: Find $f^{\prime }(x)$ for $f(x)=\frac{\sec (x)}{x^2+1}$. (Requires quotient rule)

• Finding Tangent and Normal Lines: A classic application where you must find the derivative, evaluate it at a point to find the slope, find the coordinates of the point, and write the equation of a line.

  • Example: "Write the equation of the line tangent to $g(x)=\cot (x)$ at $x=\frac{\pi }{4}$."

• Finding Locations of Horizontal Tangents: This involves setting the derivative equal to zero and solving for $x$ over a given interval.

  • Example: "Find all values of $x$ in $(0,\pi )$ for which the function $h(x)=\sqrt{3}x-2\sec (x)$ has a horizontal tangent line." (Requires solving $h^{\prime }(x)=\sqrt{3}-2\sec (x)\tan (x)=0$).

Common Mistakes

• Sign Errors: The most frequent mistake is forgetting the negative sign on the derivatives of $\cot (x)$ and $\csc (x)$.

• Formula Swapping: Confusing the derivative formulas, such as writing $\frac{d}{dx}(\sec (x))={\sec }^2(x)$ (confusing it with tangent) or $\frac{d}{dx}(\tan (x))=\sec (x)\tan (x)$ (confusing it with secant).

• Forgetting the Product/Quotient Rule: When faced with a function like $x^2\tan (x)$, a common error is to differentiate each part separately ($2x\cdot {\sec }^2(x)$) instead of correctly applying the product rule.

• Unit Circle Errors: Making arithmetic mistakes when evaluating trigonometric functions at common angles like $\frac{\pi }{6}$, $\frac{\pi }{4}$, or $\frac{\pi }{3}$. This is especially costly in tangent/normal line problems.

• Incorrectly Simplifying Identities: When deriving the formulas or simplifying a resulting derivative, students may misapply a Pythagorean or reciprocal identity, leading to an incorrect final expression.