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# Finding tangent line equations using the formal definition of a limit

AP.CALC:
CHA‑2 (EU)
,
CHA‑2.B (LO)
,
CHA‑2.B.2 (EK)
,
CHA‑2.B.3 (EK)
,
CHA‑2.B.4 (EK)
,
CHA‑2.C (LO)
,
CHA‑2.C.1 (EK)
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 f at x, equals, c (provided that limit exists):
limit, start subscript, h, \to, 0, end subscript, start fraction, f, left parenthesis, c, plus, h, right parenthesis, minus, f, left parenthesis, c, right parenthesis, divided by, h, end fraction
​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 $f(x)=x^2$f, left parenthesis, x, right parenthesis, equals, x, squared at $x=3$x, equals, 3

Step 1
What's an expression for the derivative of f, left parenthesis, x, right parenthesis, equals, x, squared at x, equals, 3?

Step 2
Evaluate the correct limit from the previous step.
f, prime, left parenthesis, 3, right parenthesis, equals

f, prime, left parenthesis, 3, right parenthesis 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 f.
Step 3
What is the point we should use for the equation of the line?
left parenthesis
comma
right parenthesis

Step 4
Complete the equation of the line tangent to the graph of f, left parenthesis, x, right parenthesis, equals, x, squared at x, equals, 3.
y, equals

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 f, left parenthesis, x, right parenthesis, equals, x, squared at x, equals, 3.
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 $g(x)=x^3$g, left parenthesis, x, right parenthesis, equals, x, cubed at $x=-1$x, equals, minus, 1

Step 1
g, prime, left parenthesis, minus, 1, right parenthesis, equals, question mark

## Example 3: Finding the equation of the line tangent to $f(x)=x^2+3$f, left parenthesis, x, right parenthesis, equals, x, squared, plus, 3 at $x=-5$x, equals, minus, 5

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?
• 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).
• 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.
• y = -10(x + 5) + 28
y = -10x - 50 + 28
y = -10x - 22
• Example 3 is bugged, yet it has been bugged for a year apparantly. Fix it already.
• Now example 3 does not accept the correct equation
• 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.
• it asks for the equation of a line. a line equation is typically of the form y=mx+b
• In example 3, why does (-5+h) get squared instead of (-5+3)?
• 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
• I don't understand step 2 of Example 2. How do they get the expanded version of (-1+h)^3?
• (-1+h) x (-1+h) = get the result then multiply it in (-1+h) again
• 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
lim[x²+2hx+h²+2x+2h-x²-2x]/h
lim[2hx+h²+2h]/h
lim[h(2x+h+2)]/h
lim 2x+h+2
2x+2