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
2. Intro to Derivatives
Tangent Lines and Derivatives
Problem 38b
Textbook Question
Find an equation of the line tangent to the graph of f at (a, f(a)) for the given value of a.
f(x) = √x+2; a=7

1
Step 1: Understand that the equation of a tangent line to the graph of a function at a point (a, f(a)) is given by the formula y - f(a) = f'(a)(x - a), where f'(a) is the derivative of the function evaluated at x = a.
Step 2: Find the derivative of the function f(x) = \sqrt{x+2}. Use the chain rule: if f(x) = (x+2)^{1/2}, then f'(x) = \frac{1}{2}(x+2)^{-1/2} \cdot 1.
Step 3: Simplify the derivative: f'(x) = \frac{1}{2\sqrt{x+2}}.
Step 4: Evaluate the derivative at x = a = 7: f'(7) = \frac{1}{2\sqrt{7+2}} = \frac{1}{2\sqrt{9}} = \frac{1}{6}.
Step 5: Use the point-slope form of the tangent line equation with f(7) = \sqrt{7+2} = 3 and f'(7) = \frac{1}{6} to write the equation: y - 3 = \frac{1}{6}(x - 7).

This video solution was recommended by our tutors as helpful for the problem above
Video duration:
7mPlay a video:
Was this helpful?
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 is determined by the derivative of the function.
Recommended video:
Slopes of Tangent Lines
Derivative
The derivative of a function at a point quantifies how the function's output changes as its input changes. It is calculated as the limit of the average rate of change of the function as the interval approaches zero. For the function f(x) = √(x + 2), the derivative will provide the slope of the tangent line at the specified point.
Recommended video:
Derivatives
Point-Slope Form
The point-slope form of a linear equation is used to express the equation of a line given a point on the line and its slope. It is written 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 the tangent line once the slope and point are known.
Recommended video:
Guided course
Slope-Intercept Form
Watch next
Master Slopes of Tangent Lines with a bite sized video explanation from Nick
Start learningRelated Videos
Related Practice