Welcome to Further Derivatives and Integrals!

In your Standard Level (SL) studies, you mastered the basics of calculus: polynomials, \( \sin x \), \( \cos x \), and the "superstar" \( e^x \). Now that you are working at the Higher Level (HL), it is time to expand your mathematical toolkit. This chapter introduces the "missing" derivatives for functions like \( \tan x \), inverse trig functions, and even logarithms with bases other than \( e \).

Think of this as moving from a basic set of tools to a professional-grade workshop. These formulas allow you to model more complex real-world phenomena, such as the path of a pendulum or the growth of a population with a specific base. Don't worry if it looks like a lot to memorize—most of these are in your Formula Booklet, but understanding how they work is the key to success!

1. New Trigonometric Derivatives

We already know that the derivative of \( \sin x \) is \( \cos x \). Now, we add the "reciprocal" trig functions and \( \tan x \) to our list. Here are the four new derivatives you need to know:

  • Tangent: \( \frac{d}{dx}(\tan x) = \sec^2 x \)
  • Secant: \( \frac{d}{dx}(\sec x) = \sec x \tan x \)
  • Cosecant: \( \frac{d}{dx}(\csc x) = -\csc x \cot x \)
  • Cotangent: \( \frac{d}{dx}(\cot x) = -\csc^2 x \)
A Quick Memory Trick:

Notice a pattern? Every "co-" function (\( \cos \), \( \csc \), \( \cot \)) has a negative derivative. If it starts with "c", its derivative starts with a minus sign! Also, notice the "partnerships": \( \tan \) and \( \sec \) usually hang out together, and \( \cot \) and \( \csc \) usually hang out together.

Quick Review: Remember that \( \sec x = \frac{1}{\cos x} \), \( \csc x = \frac{1}{\sin x} \), and \( \cot x = \frac{1}{\tan x} \). You can actually prove these derivatives using the Quotient Rule from your SL content!

2. Exponentials and Logs (Base \( a \))

In SL, we focused on base \( e \). But what if the base is \( 2 \), \( 10 \), or any positive number \( a \)?

Derivatives of \( a^x \)

The derivative of \( e^x \) is just \( e^x \). For other bases, we have to "adjust" it using the natural log of the base:
\( \frac{d}{dx}(a^x) = a^x \ln a \)

Example: If \( f(x) = 5^x \), then \( f'(x) = 5^x \ln 5 \).

Derivatives of \( \log_a x \)

Similarly, for logarithms with base \( a \):
\( \frac{d}{dx}(\log_a x) = \frac{1}{x \ln a} \)

Example: The derivative of \( \log_{10} x \) is \( \frac{1}{x \ln 10} \).

3. Inverse Trigonometric Derivatives

These are perhaps the most surprising derivatives because they don't look like trigonometry at all—they involve square roots and fractions!

  • Arcsine: \( \frac{d}{dx}(\arcsin x) = \frac{1}{\sqrt{1 - x^2}} \)
  • Arccosine: \( \frac{d}{dx}(\arccos x) = -\frac{1}{\sqrt{1 - x^2}} \)
  • Arctangent: \( \frac{d}{dx}(\arctan x) = \frac{1}{1 + x^2} \)

Note: These are valid for specific domains (for example, \( |x| < 1 \) for \( \arcsin \) and \( \arccos \)) to ensure the square root stays positive and we don't divide by zero.

4. Composites with Linear Functions \( (ax + b) \)

In the IB exams, you will rarely be asked for the derivative of just \( \tan x \). Instead, you'll see "linear composites" like \( \tan(3x + 2) \).

To handle these, we use the Chain Rule. For any function \( f(ax + b) \), the derivative is \( a \cdot f'(ax + b) \).

Integration Perspective:

This is even more important for integration! If you are integrating a function where the "inside" is \( ax + b \), you must divide by \( a \).
\( \int f(ax + b) dx = \frac{1}{a} F(ax + b) + C \)

Example: Find \( \int \sec^2(4x) dx \).
Since the derivative of \( \tan(x) \) is \( \sec^2(x) \), the integral of \( \sec^2(4x) \) is \( \frac{1}{4} \tan(4x) + C \).

5. Using Partial Fractions in Integration

Sometimes you are asked to integrate a rational function (a fraction with polynomials) that looks impossible, such as:
\( \int \frac{1}{x^2 - 1} dx \)

We don't have a direct formula for this. However, we can use Partial Fractions (from Topic 1: Number and Algebra) to split the complex fraction into two simpler ones.

Step-by-Step Process:
  1. Factor the denominator: \( x^2 - 1 = (x - 1)(x + 1) \).
  2. Split the fraction: Set \( \frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1} \).
  3. Solve for \( A \) and \( B \): Use algebraic methods to find that \( A = \frac{1}{2} \) and \( B = -\frac{1}{2} \).
  4. Integrate the parts: Now integrate \( \int (\frac{1/2}{x-1} - \frac{1/2}{x+1}) dx \).

This results in: \( \frac{1}{2} \ln|x - 1| - \frac{1}{2} \ln|x + 1| + C \).

Common Mistake: Forgetting the natural log! Remember that \( \int \frac{1}{ax + b} dx = \frac{1}{a} \ln|ax + b| + C \). This is the most common way partial fractions are used in Calculus HL.

Key Takeaways for Revision

  • Check the Formula Booklet: Most of these derivatives are provided. Your job is to recognize when to use them.
  • Watch for the "Co-": Derivatives of \( \cos \), \( \cot \), and \( \csc \) are always negative.
  • Linear Composites: When differentiating \( f(ax+b) \), multiply by \( a \). When integrating, divide by \( a \).
  • Partial Fractions: Use this technique when you see a denominator that can be factored into linear terms. It transforms a difficult fraction into easy-to-integrate \( \ln \) functions.

Note: For more advanced integration techniques like substitution or integration by parts, please refer to the chapter "Integration by substitution and by parts (HL)".