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 xThis 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 xThus, 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:
Confirm the \frac{0}{0} form by plugging in the limit value.
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).
Cancel out the common factor (x-c). (This is valid because x \neq c in the limit approaching c).
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.