Welcome back to Professor Baker’s Math Class! In this session for Spring 2025, we tackled Chapter 3: Derivatives. This chapter is often considered the heart of Calculus I, where we move past limits and start mastering the rules of differentiation. If the algebra looks intimidating, don't worry—we are going to break it down step-by-step.
1. Mastering Exponents and the Power Rule
Before applying calculus, we often need to use algebra to rewrite functions into a friendly format. As seen in our first example, we dealt with roots and fractions:
$$y = \frac{1}{\sqrt{x}} - \frac{1}{\sqrt[5]{x^3}}$$
The key here is to convert these into rational exponents before differentiating:
$$y = x^{-1/2} - x^{-3/5}$$
Once in this form, the Power Rule makes finding $y'$ straightforward.
2. The Chain Rule & Trigonometry
One of the most vital concepts in this chapter is the Chain Rule. We explored how this interacts with trigonometric functions and natural logs. A classic example from the notes involves:
$$y = \ln(\sec x)$$
By treating $u = \sec x$, we found that the derivative simplifies beautifully to:
$$y' = \tan x$$
We also looked at finding the equation of a tangent line for the function $y = 4\sin^2 x$ at the point $(\pi/6, 1)$. This required not just the Chain Rule, but also recalling values from the Unit Circle to find the slope.
3. Logarithmic Differentiation
Sometimes the Product and Quotient Rules are just too messy to use directly. This is where Logarithmic Differentiation shines. We looked at a complex rational function:
$$y = \frac{(x^2+1)^4}{(2x+1)^3(3x-1)^5}$$
Instead of a nightmare Quotient Rule scenario, we take the natural log ($\ln$) of both sides first. This allows us to use log properties to expand the right side into a sum and difference of simpler terms:
$$\ln y = 4\ln(x^2+1) - 3\ln(2x+1) - 5\ln(3x-1)$$
Differentiating implicitly gives us a much cleaner path to the answer. We also proved a theoretical identity for $f(x) = (x-a)(x-b)(x-c)$ using this same method!
4. Implicit Differentiation
Not all functions are written explicitly as $y = \dots$. In our notes, we tackled an equation involving $y$ on both sides:
$$y' + \frac{y'}{y} = y^2 + 2xyy'$$
The goal here is to group all terms containing $y'$ on one side and factor it out to solve for the derivative.
5. Applications: Finding Coefficients
Finally, we applied our skills to a system of equations problem. We were asked to find a parabola $y = ax^2 + bx + c$ that passes through $(1, 4)$ and has specific tangent slopes at $x=-1$ and $x=5$.
This required setting up a system where:
- The point $(1,4)$ satisfies the original equation.
- The derivative $y' = 2ax + b$ satisfies the slope conditions at the given $x$-values.
By solving this system, we determined the exact coefficients for $a$, $b$, and $c$.
Study Tip: Pay close attention to your algebra when simplifying negative exponents and fractions. Calculus is the new concept, but algebra is usually where the mistakes happen! Keep practicing, and I'll see you in the next class.