Section 3-6 and 3-11 Class Notes and Videos (October 18th & 20th, 2022)

Welcome to this week's update for Professor Baker's Math Class! This post contains class notes, video resources, and review materials covering sections 3-6 and 3-11. Let's dive in and master these calculus concepts together!

Class Notes: Derivatives of Logarithmic Functions (Section 3-6)

In Section 3-6, we focused on mastering the derivatives of logarithmic functions. Remember these key formulas:

  • General Logarithmic Function: $\frac{d}{dx}(\log_b x) = \frac{1}{x \ln b}$
  • Natural Logarithmic Function: $\frac{d}{dx}(\ln x) = \frac{1}{x}$

Let's look at some examples where the chain rule is needed:

  • Example 1: Find the derivative of $y = \ln(x^3 + 1)$.
    • Let $u = x^3 + 1$, then $y = \ln(u)$. So $\frac{du}{dx} = 3x^2$ and $\frac{dy}{du} = \frac{1}{u}$.
    • Using the chain rule, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{u} \cdot 3x^2 = \frac{3x^2}{x^3 + 1}$.
  • Example 2: Find the derivative of $y = \ln(\sin x)$.
    • Let $u = \sin x$, then $y = \ln(u)$. So $\frac{du}{dx} = \cos x$ and $\frac{dy}{du} = \frac{1}{u}$.
    • Using the chain rule, $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{\sin x} \cdot \cos x = \cot x$.
  • Example 3: Find the derivative of $f(x) = \sqrt{\ln x}$.
    • Let $u = \ln x$, then $f(x) = \sqrt{u}$. So $\frac{du}{dx} = \frac{1}{x}$ and $\frac{df}{du} = \frac{1}{2\sqrt{u}}$.
    • Using the chain rule, $\frac{df}{dx} = \frac{df}{du} \cdot \frac{du}{dx} = \frac{1}{2\sqrt{\ln x}} \cdot \frac{1}{x} = \frac{1}{2x\sqrt{\ln x}}$.

Class Notes: Derivatives of Inverse Trigonometric and Hyperbolic Functions (Section 3-11)

Section 3-11 introduced us to the derivatives of inverse trigonometric functions and hyperbolic functions. Here's a quick recap:

Inverse Trigonometric Functions

  • $\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}}$
  • $\frac{d}{dx}(\cos^{-1} x) = -\frac{1}{\sqrt{1-x^2}}$
  • $\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1+x^2}$
  • $\frac{d}{dx}(\csc^{-1} x) = -\frac{1}{x\sqrt{x^2-1}}$
  • $\frac{d}{dx}(\sec^{-1} x) = \frac{1}{x\sqrt{x^2-1}}$
  • $\frac{d}{dx}(\cot^{-1} x) = -\frac{1}{1+x^2}$

Hyperbolic Functions

  • $\sinh x = \frac{e^x - e^{-x}}{2}$
  • $\cosh x = \frac{e^x + e^{-x}}{2}$
  • $\tanh x = \frac{\sinh x}{\cosh x}$

Derivatives of Hyperbolic Functions

  • $\frac{d}{dx}(\sinh x) = \cosh x$
  • $\frac{d}{dx}(\cosh x) = \sinh x$
  • $\frac{d}{dx}(\tanh x) = \operatorname{sech}^2 x$

For example, if $y = \cosh(\sqrt{x})$, then $\frac{dy}{dx} = \sinh(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} = \frac{\sinh(\sqrt{x})}{2\sqrt{x}}$.

Chapter 3 Review Test and Answers

To solidify your understanding, be sure to review the Chapter 3 Test and Answers. This will give you a great sense of the material and help identify areas where you may need additional practice. Remember, practice makes perfect!

Review from 10-25 Class

Don't forget to check out the review materials from the 10-25 class, included for your convenience. Consistent review is key to mastering these concepts and acing your upcoming assessments.

Keep up the great work, and don't hesitate to reach out if you have any questions. Happy studying!