Table of contents
- 0. Functions7h 52m
- Introduction to Functions16m
- Piecewise Functions10m
- Properties of Functions9m
- Common Functions1h 8m
- Transformations5m
- Combining Functions27m
- Exponent rules32m
- Exponential Functions28m
- Logarithmic Functions24m
- Properties of Logarithms34m
- Exponential & Logarithmic Equations35m
- Introduction to Trigonometric Functions38m
- Graphs of Trigonometric Functions44m
- Trigonometric Identities47m
- Inverse Trigonometric Functions48m
- 1. Limits and Continuity2h 2m
- 2. Intro to Derivatives1h 33m
- 3. Techniques of Differentiation3h 18m
- 4. Applications of Derivatives2h 38m
- 5. Graphical Applications of Derivatives6h 2m
- 6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
- 7. Antiderivatives & Indefinite Integrals1h 26m
- 8. Definite Integrals4h 44m
- 9. Graphical Applications of Integrals2h 27m
- 10. Physics Applications of Integrals 2h 22m
3. Techniques of Differentiation
The Chain Rule
Problem 91
Textbook Question
Tangent lines Determine an equation of the line tangent to the graph of y=(x²−1)² / x³−6x−1 at the point (0,−1).

1
Step 1: Identify the function and the point of tangency. The function is \( y = \frac{(x^2 - 1)^2}{x^3 - 6x - 1} \) and the point of tangency is \((0, -1)\).
Step 2: Find the derivative of the function \( y \) with respect to \( x \) to determine the slope of the tangent line. Use the quotient rule: \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \), where \( u = (x^2 - 1)^2 \) and \( v = x^3 - 6x - 1 \).
Step 3: Calculate \( u' \) and \( v' \). For \( u = (x^2 - 1)^2 \), use the chain rule: \( u' = 2(x^2 - 1) \cdot 2x = 4x(x^2 - 1) \). For \( v = x^3 - 6x - 1 \), \( v' = 3x^2 - 6 \).
Step 4: Substitute \( u, u', v, \) and \( v' \) into the quotient rule formula to find \( y' \). Simplify the expression to find the derivative.
Step 5: Evaluate the derivative \( y' \) at \( x = 0 \) to find the slope of the tangent line at the point \((0, -1)\). Use the point-slope form of a line, \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \((x_1, y_1)\) is the point of tangency, to write the equation of the tangent line.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Tangent Line
A tangent line to a curve at a given point is a straight line that touches the curve at that point without crossing it. The slope of the tangent line represents the instantaneous rate of change of the function at that point, which can be found using the derivative of the function.
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Derivative
The derivative of a function measures how the function's output value changes as its input value changes. It is calculated as the limit of the average rate of change of the function over an interval as the interval approaches zero. For a function f(x), the derivative is denoted as f'(x) or df/dx.
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Point-Slope Form
The point-slope form of a linear equation is used to write the equation of a line when the slope and a point on the line are known. It is expressed as y - y₁ = m(x - x₁), where (x₁, y₁) is a point on the line and m is the slope. This form is particularly useful for finding the equation of a tangent line.
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Slope-Intercept Form
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