Calculus Notes: Product Rule, Quotient Rule, Higher Derivatives, Chain Rule Introduction

Derivatives: Rules and Applications

The Product Rule

• Definition: The derivative of a product of two functions, $f(x) ext{ and } g(x)$, is not the product of their derivatives. This common mistake leads to incorrect answers.

• Formula: If $h(x) = f(x)g(x)$, then its derivative, usually denoted as $h'(x)$ or $\left(\frac{d}{dx}\right)[f(x)g(x)]$, is given by:
  $f'(x)g(x) + f(x)g'(x)$
  or, as stated quickly: "the first times the derivative of the second plus the second times the derivative of the first."

• Derivation (Brief Mention): The rule is derived by adding zero (e.g., $f(x+\Delta x)g(x) - f(x+\Delta x)g(x)$) in the limit definition, allowing for rearrangement and factoring to produce the formula.

• Example (Recap from previous lecture):

  • Let $f(x) = x^4$ and $g(x) = x^3$.

  • Their product is $h(x) = x^4 \cdot x^3 = x^7$.

  • The correct derivative is $h'(x) = 7x^6$.

  • Incorrect method: Taking the product of individual derivatives ($f'(x) = 4x^3$, $g'(x) = 3x^2$). Their product is $(4x^3)(3x^2) = 12x^5$, which is incorrect.

• Application Example: Find the derivative of $f(x) = x^2 \cos x$.

  • Identify the first function ($u = x^2$) and the second function ($v = \cos x$).

  • $u' = 2x$ and $v' = -\sin x$.

  • Applying the Product Rule: $f'(x) = u \cdot v' + v \cdot u'$
    $f'(x) = (x^2)(-\sin x) + (\cos x)(2x)$
    $f'(x) = -x^2 \sin x + 2x \cos x$

  • This can be rearranged as $2x \cos x - x^2 \sin x$. An $x$ can also be factored out: $x(2 \cos x - x \sin x)$.

• Extension to Three Functions: If you have a product of three functions, $h(x) = f(x)g(x)k(x)$, its derivative is: $h'(x) = f'(x)g(x)k(x) + f(x)g'(x)k(x) + f(x)g(x)k'(x)$ This means taking the derivative of one function while keeping the others the same, and summing these combinations.

  • Example: Derivative of $x^2 \sin x \cos x$

    • Derivative of $x^2$ (other two same): $(2x)(\sin x)(\cos x)$

    • Derivative of $\sin x$ (other two same): $(x^2)(\cos x)(\cos x) = x^2 \cos^2 x$

    • Derivative of $\cos x$ (other two same): $(x^2)(\sin x)(-\sin x) = -x^2 \sin^2 x$

    • Combined: $2x \sin x \cos x + x^2 \cos^2 x - x^2 \sin^2 x$

The Quotient Rule

• Definition: Used to find the derivative of a function that is the ratio of two other functions, $f(x)/g(x)$.

• Formula: If $h(x) = \frac{f(x)}{g(x)}$, then its derivative, $h'(x)$ or $\left(\frac{d}{dx}\right)\left[\frac{f(x)}{g(x)}\right]$, is: $\frac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2}$

  • Important: Subtraction is not commutative, so the order in the numerator matters. Getting it wrong will result in an incorrect sign.

  • The rule requires that $g(x) \neq 0$.

• Mnemonic: "Low d High minus High d Low over the square of what's below." (Where 'low' refers to the denominator and 'high' to the numerator, and 'd' signifies 'derivative of'). Or simpler: "Low d High minus High d Low over Low squared."

• Application Example 1: Derivative of $\tan x$

  • Recall that $\tan x = \frac{\sin x}{ \cos x}$ (i.e., $f(x) = \sin x$, $g(x) = \cos x$).

  • $f'(x) = \cos x$ and $g'(x) = -\sin x$.

  • Applying the Quotient Rule:
    $\frac{(\cos x)(\cos x) - (\sin x)(-\sin x)}{(\cos x)^2} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x}$

  • Using the Pythagorean identity $\cos^2 x + \sin^2 x = 1$:
    $\frac{1}{\cos^2 x} = \sec^2 x$

  • Thus, the derivative of $\tan x$ is $\sec^2 x$.

• Application Example 2: Derivative of $1/x$

  • Let $f(x) = 1$ and $g(x) = x$.

  • $f'(x) = 0$ and $g'(x) = 1$.

  • Applying the Quotient Rule:
    $\frac{(x)(0) - (1)(1)}{(x)^2} = \frac{0 - 1}{x^2} = -\frac{1}{x^2}$

  • This matches the Power Rule result for $x^{-1}$ (i.e., $-1x^{-2} = -1/x^2$).

• Application Example 3: Derivative of $\frac{5x-2}{x^2+1}$

  • $f(x) = 5x-2 \implies f'(x) = 5$

  • $g(x) = x^2+1 \implies g'(x) = 2x$

  • Quotient Rule: $\frac{(x^2+1)(5) - (5x-2)(2x)}{(x^2+1)^2}$

  • Simplify numerator: $5x^2+5 - (10x^2-4x) = 5x^2+5 - 10x^2+4x = -5x^2+4x+5$

  • Result: $\frac{-5x^2+4x+5}{(x^2+1)^2}$

Derivatives of Trigonometric Functions

• Previously established: $\frac{d}{dx}(\sin x) = \cos x$ and $\frac{d}{dx}(\cos x) = -\sin x$.

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

• Other Trigonometric Derivatives:

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

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

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

• Pattern: Derivatives of 'co-functions' (cosine, cotangent, cosecant) typically include a negative sign. For example, $\frac{d}{dx}(\cos x) = -\sin x$ and $\frac{d}{dx}(\cot x) = -\csc^2 x$.

Higher-Order Derivatives

• Concept: When you take the derivative of a function, you get another function. You can then take the derivative of that new function, and so on. This process generates higher-order derivatives.

• Notation:

  • First derivative: $f'(x)$, $y'$, $\frac{dy}{dx}$, $\frac{d}{dx}[f(x)]$

  • Second derivative: $f''(x)$, $y''$, $\frac{d^2y}{dx^2}$, $\frac{d^2}{dx^2}[f(x)]$

  • Third derivative: $f'''(x)$, $y'''$, $\frac{d^3y}{dx^3}$, $\frac{d^3}{dx^3}[f(x)]$

  • Nth derivative (for $n \ge 4$): Using numbers in parentheses to avoid messy primes: $f^{(n)}(x)$, $y^{(n)}$, $\frac{d^ny}{dx^n}$ (or Roman numerals for lower orders, e.g., $y^{(iv)}$).

• Example with $\sin x$:

  • $f(x) = \sin x$

  • $f'(x) = \cos x$

  • $f''(x) = -\sin x$

  • $f'''(x) = -\cos x$

  • $f^{(4)}(x) = \sin x$

  • The pattern repeats every four derivatives. This illustrates that $f''(x) = -f(x)$ for sine and cosine functions.

• Physical Applications: Higher-order derivatives are crucial in physics and engineering:

  • If $s(t)$ is the position function of an object:

    • $s'(t) = v(t)$, which is the velocity function.

    • $s''(t) = v'(t) = a(t)$, which is the acceleration function. Acceleration is the rate of change of velocity.

    • For a falling object under constant gravity (like on Earth or the Moon), acceleration is constant (e.g., $a(t) = -9.8 \text{ m/s}^2$ on Earth, $-1.62 \text{ m/s}^2$ on the Moon).

  • Most physical applications rarely require derivatives beyond the second order.

  • Higher-order derivatives (third, fourth, etc.) are sometimes used in numerical analysis for understanding error in approximations.

Review: Evaluating Limits by Factoring Polynomials (When $0/0$ Indeterminate Form)

• Rule: If substituting the limit value into a rational function (a ratio of polynomials) results in the indeterminate form $\frac{0}{0}$, it indicates that further work (e.g., factoring) is needed to find the limit.

• Key Fact: If a polynomial $P(x)$ evaluates to 0 when $x=c$ (i.e., $P(c)=0$), then $(x-c)$ is a factor of $P(x)$.

• Strategy: For $\frac{0}{0}$ forms with polynomials:

  1. Confirm the $\frac{0}{0}$ form by plugging in the limit value.

  2. Factor out the common factor $(x-c)$ (where $c$ is the limit value) from both the numerator and the denominator.

     • For quadratics, this is straightforward.

     • For cubics or higher-degree polynomials, if you know $(x-c)$ is a factor, you can use:

       • Reasoning/Inspection: For $x^3+3x-4$ as $x \to 1$, we know $(x-1)$ is a factor. We need $(x-1)(ax^2+bx+d)$. Matching coefficients:

         • $x \cdot ax^2 = x^3 \implies a=1$.

         • $-1 \cdot d = -4 \implies d=4$.

         • Now we have $(x-1)(x^2+bx+4)$. Expand: $x^3+bx^2+4x-x^2-bx-4 = x^3+(b-1)x^2+(4-b)x-4$.

         • Comparing with $x^3+3x-4$, we need $(b-1)x^2 = 0x^2 \implies b=1$. Also, $(4-b)x = 3x \implies 4-1 = 3$, which means $b=1$ works.

         • So, $x^3+3x-4 = (x-1)(x^2+x+4)$.

       • Polynomial Long Division: Divide the polynomial by $(x-c)$.

  3. Cancel out the common factor $(x-c)$. (This is valid because $x \neq c$ in the limit approaching $c$).

  4. Plug the limit value into the simplified expression to find the limit.

• Example 1: Find the limit as $x \to 1$ of $\frac{x^2+5x-6}{x^2-3x+2}$.

  • Plugging in $x=1$ gives $\frac{1+5-6}{1-3+2} = \frac{0}{0}$. Do more work.

  • Factor: Numerator $(x-1)(x+6)$; Denominator $(x-1)(x-2)$.

  • Expression becomes $\frac{(x-1)(x+6)}{(x-1)(x-2)}$.

  • Cancel $(x-1)$: $\frac{x+6}{x-2}$.

  • Plug in $x=1$: $\frac{1+6}{1-2} = \frac{7}{-1} = -7$.

• Example 2: Find the limit as $x \to 1$ of $\frac{x^3+3x-4}{x^2+4x-5}$.

  • Plugging in $x=1$ gives $\frac{1+3-4}{1+4-5} = \frac{0}{0}$. Do more work.

  • Factor: Denominator $(x-1)(x+5)$.

  • Factor Numerator (as shown above): $(x-1)(x^2+x+4)$.

  • Expression becomes $\frac{(x-1)(x^2+x+4)}{(x-1)(x+5)}$.

  • Cancel $(x-1)$: $\frac{x^2+x+4}{x+5}$.

  • Plug in $x=1$: $\frac{1^2+1+4}{1+5} = \frac{6}{6} = 1$.

• DNE (Does Not Exist) Cases: If you get $\frac{\text{number}}{0}$ (where the number is non-zero) after simplification, the limit DNE.

The Chain Rule (Introduction)

• Concept: Used to find the derivative of a composite function, i.e., a function within a function.

• Formula: If $y = f(g(x))$ (read as "f of g of x"), then its derivative is: $y' = f'(g(x)) \cdot g'(x)$

  • In simpler terms: "Take the derivative of the 'whole thing' (the outer function with the inner function unchanged), and then multiply by the derivative of the 'inside' (the inner function)."

• Application Example: Find the derivative of $f(x) = (x^3+1)^2$.

  • Identify the outer function: $f_{outer}(u) = u^2$ (where $u = x^3+1$).

  • Identify the inner function: $g(x) = x^3+1$.

  • Derivative of the outer function with inner function unchanged: $f'_{outer}(u) = 2u$, so $2(x^3+1)$.

  • Derivative of the inner function: $g'(x) = 3x^2$.

  • Applying Chain Rule: $f'(x) = 2(x^3+1) \cdot (3x^2) = 6x^2(x^3+1)$.

• Verification by Expansion: Expand $f(x) = (x^3+1)^2 = x^6 + 2x^3 + 1$.

  • Take the derivative: $f'(x) = 6x^5 + 6x^2 + 0 = 6x^5 + 6x^2$.

  • Expanding the Chain Rule result: $6x^2(x^3+1) = 6x^5 + 6x^2$.

  • The results match, confirming the Chain Rule.

• Multiple Links: For more complex nested functions (e.g., $f(g(h(x)))$), the chain rule can be applied successively (e.g., $f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)$).

Test Preparation Notes

• Test 1: Scheduled for September 30th.

• Content Distribution: The test will likely feature an equal distribution of topics covered.

  • Two problems will be high-value with multiple sub-parts.

  • One page (approx. 6 questions) will focus on limits, including problems like the factoring examples discussed.

  • Another page (approx. 3 questions) will involve various derivative calculations.

  • One question on the definition of a derivative.

  • One question on continuity.

  • Problem 2 will be on pre-calculus topics (inequalities: absolute value, quadratic).

  • Problem 1 will cover implicit differentiation, which will be introduced closer to the test date (e.g., this coming Friday).

  • The test covers precalculus, limits, and derivatives in order, concluding with an application-oriented problem.