Rules for derivatives (Calc 1; Lecture 1-8; Fall 26)

Overview

Calculus I differentiation rules replace repeated limit calculations with reusable building blocks: the constant, power, and exponential rules, plus linearity, product, reciprocal, and quotient rules. Beard Meets Calculus shows how to derive or apply these rules, use them to solve tangent-slope problems, and handle a complicated derivative by working from its outermost operation inward.

Key takeaways

To make a piecewise function differentiable at a join, enforce both continuity and equal one-sided derivatives; for the example at x = 1, the resulting system gives a = 5 and b = -4.
The power rule converts d(x^k)/dx to kx^(k-1); rewriting roots and reciprocals as powers lets the same rule handle √x, 1/x^4, and x^e.
The derivative of e^x is e^x, while the derivative of e^100 with respect to x is zero because e^100 is constant.
The product rule is (fg)' = f'g + fg', not f'g'; for three factors, each term differentiates exactly one factor.
A horizontal tangent occurs where the derivative equals zero; for y = x^3 - 12x + 17, this gives x = ±2.
For complicated derivatives, identify the outermost operation first, then apply its rule and work inward; a quotient containing x^2e^x requires both the quotient and product rules.

Chapters

0:00 Quickfire Review and Matching Piecewise Derivatives at x = 1

The opening review checks arithmetic, solves 2|x| = x + 2 as x = -2/3 or x = 2, and evaluates a limit at infinity by dividing by the fastest-growing e^x terms.
For a function defined by 2x^3 - 1 below x = 1 and ax^2 + bx above it, differentiability requires both continuity and matching one-sided slopes.
Continuity gives a + b = 1; matching derivatives gives 2a + b = 6, so a = 5 and b = -4.
At a piecewise join, derivative formulas initially apply only on their respective open intervals; the joining point must be checked separately.

9:03 Why Differentiation Rules Replace Repeated Limit Work

The main goal is to make derivatives easier to compute without directly repeating the limit definition for every function.
Limits remain the underlying machinery, but rules provide a more efficient way to differentiate increasingly complicated expressions.
The rules are organized as function building blocks and combination rules, and students are encouraged to learn and practice each one.

11:29 Constant, Power, and Exponential Rules from Limits

A constant such as 1 has derivative 0 because its graph is horizontal and its value does not change.
For a positive whole-number exponent k, factoring z^k - x^k cancels the z - x denominator in the limit definition; the remaining k terms each become x^(k-1), giving d(x^k)/dx = kx^(k-1).
The derivative of e^x is e^x: factoring e^x from the difference quotient leaves the defining limit (e^h - 1)/h, which tends to 1.
The number e is introduced as approximately 2.71828 and connected to the limit (1 + h)^(1/h) as h approaches zero.

21:32 Applying Power Rules to Constants, Roots, and Reciprocals

The derivative of e^100 with respect to x is 0 because e^100 contains no x and is therefore a constant.
Rewrite √x as x^(1/2) before differentiating; the power rule gives (1/2)x^(-1/2), or 1/(2√x).
Rewrite 1/x^4 as x^(-4); its derivative is -4x^(-5), equivalently -4/x^5.
The exponent can be any number in the power rule: d(x^e)/dx = e x^(e-1), which is distinct from the exponential function e^x.

25:57 Linearity Breaks Sums into Differentiable Pieces

Linearity allows derivatives to distribute across addition and subtraction, while constant factors can be pulled outside the derivative.
For a multi-term expression involving powers, 1/x, e^x, and a constant such as π^3, differentiate each term separately and use the relevant building-block rule.
Rewriting expressions such as 1/x as x^(-1) makes them compatible with the power rule.
The rationale is that limits themselves distribute over sums and allow constant factors to be pulled out.

30:06 Horizontal Tangents Become a Derivative-Zero Equation

A horizontal tangent has slope zero, and the derivative gives the tangent slope at each point.
For y = x^3 - 12x + 17, linearity and the power rule give y' = 3x^2 - 12.
Solving 3x^2 - 12 = 0 yields x^2 = 4, so horizontal tangents occur at x = -2 and x = 2.
The derivative rules turn a problem that would require a lengthy limit calculation into a short algebraic equation.

32:06 Product Rule: Differentiate One Factor at a Time

For f(x)g(x), the product rule is (fg)' = f'g + fg'; differentiate one factor in each term while leaving the other unchanged.
Differentiating both factors simultaneously is incorrect; the rule gives each factor a separate turn.
The limit proof adds and subtracts an intermediate product so the change in f and the change in g can be separated.
An area diagram offers another interpretation: the first-order change comes from two side strips, while the small corner formed by both changes is negligible.

37:36 Extending the Product Rule to Three Factors and e^x

For f(x)g(x)h(x), grouping two factors and applying the product rule twice produces f'gh + fg'h + fgh'.
The three-factor result follows the same pattern: differentiate one factor per term and leave the other two unchanged.
For x^(1/3)e^x, the product rule gives (1/3)x^(-2/3)e^x + x^(1/3)e^x.
When factoring the result, e^x and the smaller exponent x^(-2/3) are common factors; the expression becomes e^x x^(-2/3)(1/3 + x).

44:21 Reciprocal and Quotient Rules for Fractions

For a reciprocal, d(1/g(x))/dx = -g'(x)/[g(x)]^2.
Combining the reciprocal and product rules gives the quotient rule: (f/g)' = [f'g - fg']/(g^2).
The mnemonic “low d high minus high d low over low squared” preserves the order of the numerator terms and the squared denominator.
The minus sign in f'g - fg' is a common test point, so the quotient rule needs careful practice.

46:26 Differentiate a Layered Expression from the Outermost Rule Inward

For a quotient with numerator x^(1/3) - 7x^4 and denominator 8 + x^2e^x, begin with the quotient rule because division is the outermost operation.
Differentiate the numerator by linearity and the power rule: (1/3)x^(-2/3) - 28x^3.
Differentiate the denominator by linearity and the product rule: the constant 8 contributes 0, while (x^2e^x)' = 2xe^x + x^2e^x.
The final derivative places denominator times numerator derivative minus numerator times denominator derivative over (8 + x^2e^x)^2; the recommended method is to handle one rule-layer at a time.

Summary, takeaways, and chapters were generated by AI from the video's transcript and may contain errors. The video belongs to its creator, Beard Meets Calculus.