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).

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

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

Well, I answered, y=-10x+22, but it said wrong. But this is correct too, right?

Sign error. It should be y = -10x - 22, not y=-10x + 22

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

Can we replace y with the x^2 +3 in the answer above?

Nope, you can't. x^2 + 3 is f(x), which is the equation of the *curve*. y is −10(x+5)+28, which is the equation of the *line* tangent to the curve. Clearly, they aren't the same thing and hence, cannot be equated to each other.

For question 1: what is step 4? How should I even interpret the question? This is so frustrating. For question 2: Why do we use the large formula? What is indicating that we have a g(t+h) function instead of a g(x)-g(t) function? And then I am asked to choose a point on the line after finding out what the slope of the tangent is. Is this darts?

Step 4 is just asking you to use all the information you already solved for (the slope and point) to get the equation of the tangent. You must already know that a unique line can be defined by a slope and a single point, so that's what they want you to do. You could use that formula too. With it, you'll get lim (x --> -1) (x^(3)+1)/(x+1). Expanding the numerator with the identity (a^(3) + b^(3) = (a+b)(a^(2) - ab + b^(2)), we get lim (x --> -1) (x+1)(x^(2) - 2x + 1)/(x+1). The x+1's cancel out, and you're left with lim (x --> -1) (x^(2) - 2x + 1). Substitute x = -1, and we get 3, which is exactly what we got with the other formula too. I reckon they used the other one because (a ± b)^(3) is more easy to calculate (incase you forget the identity) than a^(3) ± b^(3) (for which deriving the identity requires a but more effort) You could go the other way too, honestly. You can first find a point on the line and then find the slope. The exercise just wants to familiarize you with the idea and give an algorithm for students to follow. You are more than welcome to experiment and do stuff in a different way, as long as you obey all the rules of maths.

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

Main content

Course: Class 11 Math (India) - Hindi > Unit 8

Lesson 2: Derivation definition

Finding tangent line equations using the formal definition of a limit

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

x = c

(provided that limit exists):

lim_{h \to 0} \frac{f (c + h) - f (c)}{h}

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 $f (x) = x^{2}$ ‍ at $x = 3$ ‍

Step 1

What's an expression for the derivative of $f (x) = x^{2}$ ‍ at $x = 3$ ‍?

Step 2

Evaluate the correct limit from the previous step.

f^{'} (3) =

f^{'} (3)

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?

(

,

)

Step 4

Complete the equation of the line tangent to the graph of $f (x) = x^{2}$ ‍ at $x = 3$ ‍.

y =

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 (x) = x^{2}

x = 3

Example 2: Finding the equation of the line tangent to the graph of $g (x) = x^{3}$ ‍ at $x = - 1$ ‍

Step 1

g^{'} (- 1) = ?

Example 3: Finding the equation of the line tangent to $f (x) = x^{2} + 3$ ‍ at $x = - 5$ ‍

Let's do this one without all the steps.

What is the equation of the tangent line?

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BB8FN2187
Posted 6 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
(20 votes)
Answer
- ( ͡° ͜ʖ ͡°)
  Posted 6 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...”
  (134 votes)
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 = -...”
(34 votes)
Answer
djschrock00
Posted 6 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
Ivan Korniienko
Posted 3 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...”
(19 votes)
Answer
- henrydavidthomas
  Posted 3 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
  (16 votes)
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 ...”
(13 votes)
Answer
- redfoer
  Posted 4 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...”
  (5 votes)
Keiko Whitlock
Posted 10 months ago. Direct link to Keiko Whitlock's post “Well, I answered, y=-10x+...”
Well, I answered, y=-10x+22, but it said wrong. But this is correct too, right?
Button navigates to signup pageComment on Keiko Whitlock's post “Well, I answered, y=-10x+...”
(3 votes)
Answer
- Venkata
  Posted 10 months ago. Direct link to Venkata's post “Sign error. It should be ...”
  Sign error. It should be y = -10x - 22, not y=-10x + 22
  Comment on Venkata's post “Sign error. It should be ...”
  (9 votes)
Vector Inc.
Posted 3 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 pageComment on Vector Inc.'s post “I don't understand step 2...”
(3 votes)
Answer
- ahmada timo
  Posted 3 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)
avs68325
Posted 7 months ago. Direct link to avs68325's post “For question 1: what is s...”
For question 1: what is step 4? How should I even interpret the question? This is so frustrating.

For question 2: Why do we use the large formula? What is indicating that we have a g(t+h) function instead of a g(x)-g(t) function?

And then I am asked to choose a point on the line after finding out what the slope of the tangent is. Is this darts?
Button navigates to signup pageButton navigates to signup page
(2 votes)
Answer
- Venkata
  Posted 7 months ago. Direct link to Venkata's post “Step 4 is just asking you...”
  Step 4 is just asking you to use all the information you already solved for (the slope and point) to get the equation of the tangent. You must already know that a unique line can be defined by a slope and a single point, so that's what they want you to do.
  
  You could use that formula too. With it, you'll get lim (x --> -1) (x^(3)+1)/(x+1). Expanding the numerator with the identity (a^(3) + b^(3) = (a+b)(a^(2) - ab + b^(2)), we get lim (x --> -1) (x+1)(x^(2) - 2x + 1)/(x+1). The x+1's cancel out, and you're left with lim (x --> -1) (x^(2) - 2x + 1). Substitute x = -1, and we get 3, which is exactly what we got with the other formula too.
  
  I reckon they used the other one because (a ± b)^(3) is more easy to calculate (incase you forget the identity) than a^(3) ± b^(3) (for which deriving the identity requires a but more effort)
  
  You could go the other way too, honestly. You can first find a point on the line and then find the slope. The exercise just wants to familiarize you with the idea and give an algorithm for students to follow. You are more than welcome to experiment and do stuff in a different way, as long as you obey all the rules of maths.
  Comment on Venkata's post “Step 4 is just asking you...”
  (5 votes)
sufyan.desai05
Posted 5 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 5 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)
aditi.91.gupta
Posted a year ago. Direct link to aditi.91.gupta's post “Can we replace y with the...”
Can we replace y with the x^2 +3 in the answer above?
Button navigates to signup pageButton navigates to signup page
(3 votes)
Answer
- Venkata
  Posted a year ago. Direct link to Venkata's post “Nope, you can't. x^2 + 3 ...”
  Nope, you can't. x^2 + 3 is f(x), which is the equation of the curve. y is −10(x+5)+28, which is the equation of the line tangent to the curve. Clearly, they aren't the same thing and hence, cannot be equated to each other.
  Button navigates to signup page
  (3 votes)

Class 11 Math (India) - Hindi

Course: Class 11 Math (India) - Hindi > Unit 8

Example 1: Finding the equation of the line tangent to the graph of f(x)=x2‍ at x=3‍

Example 2: Finding the equation of the line tangent to the graph of g(x)=x3‍ at x=−1‍

Example 3: Finding the equation of the line tangent to f(x)=x2+3‍ at x=−5‍

Want to join the conversation?

Example 1: Finding the equation of the line tangent to the graph of $f (x) = x^{2}$ ‍ at $x = 3$ ‍

Example 2: Finding the equation of the line tangent to the graph of $g (x) = x^{3}$ ‍ at $x = - 1$ ‍

Example 3: Finding the equation of the line tangent to $f (x) = x^{2} + 3$ ‍ at $x = - 5$ ‍