AP®︎/College Calculus AB
Course: AP®︎/College Calculus AB > Unit 2Lesson 2: Defining the derivative of a function and using derivative notation
- Formal definition of the derivative as a limit
- Formal and alternate form of the derivative
- Worked example: Derivative as a limit
- Worked example: Derivative from limit expression
- Derivative as a limit
- The derivative of x² at x=3 using the formal definition
- The derivative of x² at any point using the formal definition
- Finding tangent line equations using the formal definition of a limit
Finding tangent line equations using the formal definition of a limit
This structured practice takes you through three examples of finding the equation of the line tangent to a curve at a specific point.
We can calculate the slope of a tangent line using the definition of the derivative of a function at (provided that limit exists):
Once we've got the slope, we can find the equation of the line. This article walks through three examples.
Function f is graphed. The positive x-axis includes value c. The graph is a curve. The curve starts in quadrant 2, moves downward to a point in quadrant 1, moves upward through a point at x = c, and ends in quadrant 1. A tangent line starts in quadrant 4, moves upward, touches the curve at the point at x = c, and ends in quadrant 1.
Example 1: Finding the equation of the line tangent to the graph of at
What's an expression for the derivative of at ?
Evaluate the correct limit from the previous step.
gives us the slope of the tangent line. To find the complete equation, we need a point the line goes through.
Usually, that point will be the point where the tangent line touches the graph of .
What is the point we should use for the equation of the line?
Complete the equation of the line tangent to the graph of at .
And we're done! Using the definition of the derivative, we were able to find the equation for the line tangent to the graph of at .
Function f is graphed. The x-axis goes from negative 12 to 12. The graph is a U-shaped curve. The curve starts in quadrant 2, moves downward to (0, 0), moves upward through a point at about (3, 9), and ends in quadrant 1. A tangent line starts in quadrant 4, moves upward, touches the curve at the point, and ends in quadrant 1.
Example 2: Finding the equation of the line tangent to the graph of at
Example 3: Finding the equation of the line tangent to at
Let's do this one without all the steps.
What is the equation of the tangent line?
Want to join the conversation?
- Why do we need to know this? Couldn't we just take the derivative using derivative rules and be faster/easier?(17 votes)
- Yes, derivative rules are faster and easier than using the "first principles" definition of the derivative (with limits and the definition of slope), but the only reason we have those rules is because of that definition of the derivative. In practice, you do use the derivative rules, not the first principles definition, to find derivatives, but it's really important to understand the intuition of why we can even find instantaneous slope in the first place (which was a nonsensical idea to people more recently than you would think).(99 votes)
- I found using point-slope form to solve for the final equation to be easier than this. I entered my answer to the third question as -10x-22 and that is correct (if you simplify the answer given, it is the same) but it might be confusing to others.(29 votes)
- y = -10(x + 5) + 28
y = -10x - 50 + 28
y = -10x - 22(24 votes)
- Example 3 is bugged, yet it has been bugged for a year apparantly. Fix it already.(10 votes)
- Now example 3 does not accept the correct equation(4 votes)
- Deat Khan Academy, please communicate it clearer that the "Example 3: Finding the equation of the line tangent" MUST contain "y=" at the front of the answer.
I nearly lost my mind trying to understand where was my arithmetic wrong.(4 votes)
- it asks for the equation of a line. a line equation is typically of the form y=mx+b(9 votes)
- I don't understand step 2 of Example 2. How do they get the expanded version of (-1+h)^3?(3 votes)
- (-1+h) x (-1+h) = get the result then multiply it in (-1+h) again(4 votes)
- In example 3, why does (-5+h) get squared instead of (-5+3)?(2 votes)
- The function is f(x)=x^2+3 (For each value of x you gonna to square it then plus 3)
And so for f(-5+h), you have to square (-5+h) first and then plus 3
Another way to say it, to calculate f(-5+h), whenever you see x in the function, you replace x with -5+h, and you'll get f(-5+h)=(-5+h)^2+3
That's why (-5+h) get squared(5 votes)
- what is the derivative of x^2 + 2x using the limit definition(1 vote)
- You have the limit as h->0 of [(x+h)²+2(x+h)-[x²+2x]]/h
- In step 2 explanation of the first example why was there a 6h added to the equation?(1 vote)
- (a+b)^2 = a^2 + 2ab+ b^2
(3+h)^2 = 9 + 2.3.h + h^2
= 9 + 6h+ h^2(2 votes)
- why is the slope of (3,9) = 6? Shouldn't it be 3?(1 vote)
- You cannot prove something that is not true. I would double check your working out.(2 votes)