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).

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

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

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

what is the derivative of x^2 + 2x using the limit definition

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

In step 2 explanation of the first example why was there a 6h added to the equation?

(a+b)^2 = a^2 + 2ab+ b^2 (3+h)^2 = 9 + 2.3.h + h^2 = 9 + 6h+ h^2

why is the slope of (3,9) = 6? Shouldn't it be 3?

You cannot prove something that is not true. I would double check your working out.

Main content

AP®︎/College Calculus AB

Course: AP®︎/College Calculus AB > Unit 2

Lesson 2: Defining the derivative of a function and using derivative notation

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)

Google Classroom

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.

Example 1: Finding the equation of the line tangent to the graph of f, left parenthesis, x, right parenthesis, equals, x, squared at 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?

(Choice A)
limit, start subscript, h, \to, 0, end subscript, start fraction, left parenthesis, 3, plus, h, right parenthesis, squared, minus, 3, squared, divided by, h, end fraction
(Choice B)
limit, start subscript, h, \to, 0, end subscript, start fraction, left parenthesis, 3, plus, h, right parenthesis, squared, minus, h, squared, divided by, h, end fraction
(Choice C)
limit, start subscript, h, \to, 0, end subscript, start fraction, 3, squared, minus, h, squared, divided by, h, end fraction
(Choice D)
limit, start subscript, h, \to, 0, end subscript, start fraction, left parenthesis, x, plus, h, right parenthesis, squared, minus, x, squared, divided by, h, end fraction

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.

Example 2: Finding the equation of the line tangent to the graph of g, left parenthesis, x, right parenthesis, equals, x, cubed at x, equals, minus, 1

Step 1

Current

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

(Choice A)
limit, start subscript, h, \to, 0, end subscript, start fraction, left parenthesis, minus, 1, plus, h, right parenthesis, cubed, minus, left parenthesis, minus, 1, right parenthesis, cubed, divided by, h, end fraction
(Choice B)
limit, start subscript, h, \to, 0, end subscript, start fraction, left parenthesis, minus, 1, right parenthesis, cubed, minus, h, cubed, divided by, h, end fraction
(Choice C)
limit, start subscript, h, \to, 0, end subscript, start fraction, left parenthesis, minus, 1, plus, h, right parenthesis, cubed, minus, h, cubed, divided by, h, end fraction
(Choice D)
limit, start subscript, h, \to, 0, end subscript, start fraction, h, cubed, minus, left parenthesis, minus, 1, right parenthesis, cubed, divided by, h, end fraction

Example 3: Finding the equation of the line tangent to f, left parenthesis, x, right parenthesis, equals, x, squared, plus, 3 at x, equals, minus, 5

Let's do this one without all the steps.

What is the equation of the tangent line?

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BB8FN2187
Posted 5 years ago. Direct link to BB8FN2187's post “Why do we need to know th...”
Why do we need to know this? Couldn't we just take the derivative using derivative rules and be faster/easier?
Button navigates to signup pageButton navigates to signup page
(17 votes)
Answer
- ( ͡° ͜ʖ ͡°)
  Posted 5 years ago. Direct link to ( ͡° ͜ʖ ͡°)'s post “Yes, derivative rules are...”
  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).
  Comment on ( ͡° ͜ʖ ͡°)'s post “Yes, derivative rules are...”
  (109 votes)
djschrock00
Posted 5 years ago. Direct link to djschrock00's post “I found using point-slope...”
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.
Button navigates to signup pageComment on djschrock00's post “I found using point-slope...”
(29 votes)
Answer
Abel Romo
Posted 4 years ago. Direct link to Abel Romo's post “y = -10(x + 5) + 28 y = -...”
y = -10(x + 5) + 28
y = -10x - 50 + 28
y = -10x - 22
Button navigates to signup pageComment on Abel Romo's post “y = -10(x + 5) + 28 y = -...”
(24 votes)
Answer
Mihrr
Posted 5 years ago. Direct link to Mihrr's post “Example 3 is bugged, yet ...”
Example 3 is bugged, yet it has been bugged for a year apparantly. Fix it already.
Button navigates to signup pageComment on Mihrr's post “Example 3 is bugged, yet ...”
(10 votes)
Answer
- redfoer
  Posted 3 years ago. Direct link to redfoer's post “Now example 3 does not ac...”
  Now example 3 does not accept the correct equation
  Comment on redfoer's post “Now example 3 does not ac...”
  (4 votes)
Ivan Korniienko
Posted 2 years ago. Direct link to Ivan Korniienko's post “Deat Khan Academy, please...”
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.
Button navigates to signup pageComment on Ivan Korniienko's post “Deat Khan Academy, please...”
(5 votes)
Answer
- henrydavidthomas
  Posted 2 years ago. Direct link to henrydavidthomas's post “it asks for the equation ...”
  it asks for the equation of a line. a line equation is typically of the form y=mx+b
  Button navigates to signup page
  (11 votes)
Vector Inc.
Posted 2 years ago. Direct link to Vector Inc.'s post “I don't understand step 2...”
I don't understand step 2 of Example 2. How do they get the expanded version of (-1+h)^3?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- ahmada timo
  Posted 2 years ago. Direct link to ahmada timo's post “(-1+h) x (-1+h) = get the...”
  (-1+h) x (-1+h) = get the result then multiply it in (-1+h) again
  Button navigates to signup page
  (4 votes)
sufyan.desai05
Posted 4 years ago. Direct link to sufyan.desai05's post “In example 3, why does (-...”
In example 3, why does (-5+h) get squared instead of (-5+3)?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Kiên P.S.
  Posted 4 years ago. Direct link to Kiên P.S.'s post “The function is f(x)=x^2+...”
  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
  Comment on Kiên P.S.'s post “The function is f(x)=x^2+...”
  (5 votes)
Howard Slocum
Posted 3 years ago. Direct link to Howard Slocum's post “what is the derivative of...”
what is the derivative of x^2 + 2x using the limit definition
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(1 vote)
Answer
- kubleeka
  Posted 3 years ago. Direct link to kubleeka's post “You have the limit as h->...”
  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
  Button navigates to signup page
  (3 votes)
Micah270
Posted 3 years ago. Direct link to Micah270's post “In step 2 explanation of ...”
In step 2 explanation of the first example why was there a 6h added to the equation?
Button navigates to signup pageComment on Micah270's post “In step 2 explanation of ...”
(1 vote)
Answer
- Anil
  Posted 3 years ago. Direct link to Anil's post “(a+b)^2 = a^2 + 2ab+ b^2 ...”
  (a+b)^2 = a^2 + 2ab+ b^2
  (3+h)^2 = 9 + 2.3.h + h^2
  = 9 + 6h+ h^2
  Button navigates to signup page
  (2 votes)
石乐志大师
Posted a year ago. Direct link to 石乐志大师's post “why is the slope of (3,9)...”
why is the slope of (3,9) = 6? Shouldn't it be 3?
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(1 vote)
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- cossine
  Posted a year ago. Direct link to cossine's post “You cannot prove somethin...”
  You cannot prove something that is not true. I would double check your working out.
  Button navigates to signup page
  (2 votes)

AP®︎/College Calculus AB

Course: AP®︎/College Calculus AB > Unit 2

Example 1: Finding the equation of the line tangent to the graph of f(x)=x2f(x)=x^2f(x)=x2f, left parenthesis, x, right parenthesis, equals, x, squared at x=3x=3x=3x, equals, 3

Example 2: Finding the equation of the line tangent to the graph of g(x)=x3g(x)=x^3g(x)=x3g, left parenthesis, x, right parenthesis, equals, x, cubed at x=−1x=-1x=−1x, equals, minus, 1

Example 3: Finding the equation of the line tangent to f(x)=x2+3f(x)=x^2+3f(x)=x2+3f, left parenthesis, x, right parenthesis, equals, x, squared, plus, 3 at x=−5x=-5x=−5x, equals, minus, 5

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Example 1: Finding the equation of the line tangent to the graph of f, left parenthesis, x, right parenthesis, equals, x, squared at x, equals, 3

Example 2: Finding the equation of the line tangent to the graph of g, left parenthesis, x, right parenthesis, equals, x, cubed at x, equals, minus, 1

Example 3: Finding the equation of the line tangent to f, left parenthesis, x, right parenthesis, equals, x, squared, plus, 3 at x, equals, minus, 5