Struggling with calculus =/

This is great advice. I got all the way through Differential Equations using this method, by basically reproducing the text book, this same way. Math, while conceptual in nature, is incredibly visual when you are first learning it. If you can map out the concepts you are learning by your own hand, you will gain a greater understanding for what it is you're trying to accomplish.

 

https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/defderdirectory/DefDer.html

Limit definition for derivatives is finding the derivatives using a special formula. The proof of it is created through a simple process, but that's not important right now.

Anyway, limit definition is

[f(x+h)-f(x)]/h

From there, you just plug in for everywhere f(x) is. If there's an f(x+h), you simply add an h wherever you see an x.

For example, f(x) = 2x^2-1

So start with the numerator

f(x+h) -----> 2(x+h)^2 - 1
f(x)--------> 2x^2 - 1

h is a variable, so you don't treat it as a number until you replace it with the limit towards the end of the equation. So lets plug that all in

f(x+h) - f(x) / h
=
[2(x+h)^2 - 1] - [2x^2 - 1] / h

from there, you just use basic algebra.

[2(x+h)(x+h) - 1] - (2x^2 - 1) / h

2x^2 +2xh +2xh + 2h^2 - 1 - (2x^2 -1) / h

Cancel out what you can, now.

4xh + 2h^2 / h

Divide by 'h' to cancel out from the numerator.

4x + 2h <----------- (I believe this is what they're asking of you, when they ask for the "equation". It might just be 4x, though.)

h > 0 (h approaches 0, so you plug in)

= 4x + 2(0) = 4x <----------- Another possible answer


plug in (0, -1)

4(0)

= 0







******Anyone feel free to correct any mistakes, please. Kind of rushed this for the sake of him.

 

Looks like New Generation already answered how to compute the derivative using the limits definition. Here's how I'd answer it:

Start by plotting the graph of f(x) on a sheet of paper, for x = -5, -4, -3, ..., 3, 4, 5. Then, draw a line that is tangent to the curve you just drew and intersects it at (0, -1). Your objective is to find the equation for this line (if you drew the curve it should be readily obvious, but we'll work through the math anyway). The equation will be of the form:

y = mx + b, where m is the slope of the line.

Since the line crosses the point (0, -1), we know that b = -1. So then we have the equation:

y = mx - 1

The slope of this line is equal to the derivative of the function you drew, at x=0. You can therefore use the limit definition of the derivative to calculate this slope:

f'(x) = lim (h -> 0) [ f( x + h) - f(x) ] / h, where

f(x) = 2x^2 - 1

Take h to be a very small number. That would result in an approximation of f'(x) at x=0 being:

Code:

f'(0) ~= (f(0+h) - f(0)) /  h = ((2*(h)^2 - 1) -  (2*(0)^2 - 1)) / h
       = 2*h^2 / h
       = 2*h

So, for the limit h approaches 0, f'(0) approaches 0. In other words, the slope of our tangent line (m) at the point (0, -1) must be 0.

So, final answer should be y = -1.

 

do u even own a ti-83 brah?

 

Edit. Above 2 posters had the same thing as me first

Just adding on, after you get the slope of the tangent line to be =0, you can also use:

y-y1=m(x-x1) to find the equation, which is what I find easier.

y-(-1)=0(x-0)
y+1=0
y=-1

 

Should have kept yours, IMO, because your approach was a little different. It might have been helpful for him to see that there's different ways to solve it.

 

I thought new generation's was better organized but here's mine anyways.

 

The answer is y = -1. Read through my explanation and see if that helps.

 

New generation posted how to find the slope of the tangent line, which is the same thing as the derivative of the given equation.
4x is the slope of the tangent line. In order to solve for the slope, they gave you the points x=0,y=-1. You plug in 0 for x to get slope (m)= 4x=0.

Then you plug into the equation y-y1=m(x-x1) to get the actual equation of the tangent line.

y+1=0(x-0)

y=-1.

 

Yea, sorry I couldn't really figure out what to do with the point. It's been awhile since I've taken Calc, and I'm just now re-taking it in college. Anyway, if we were to solve this derivative the easy way, the equation would be 4x. That's also what you get using the limit method.

So yes, I THINK the answer is 4x. Then again, I could be wrong. The above posters probably know the material much better than I do, and understand the question more.

 

EDIT: Never mind. Everyone already did it lol and I didn't use the limit definition.

 

Yes, to find the equation of the tangent line, you can use point slope formula.
They gave you a point, and you find the slope with the derivative. Then you just fill in the blanks into the point slope form to find the equation.

 

RedRed, Use the limit definition

But yeah, to make you feel better op, this gets much easier once you learn power/product/quotient rule, which I would guess is what you will learn next. Using the definition to find the derivative is much longer than necessary. You just need to get this down for free response stuff. Multiple choice can always be done with the rules.

This might help-----
Just remember when you see this type of problem to find the eqn of a tangent line:
1. The first thing you should think of is y-y1=m(x-x1). That is what you are shooting for as your final answer.
2. Fill in the given point into the point slope equation. In this case, (y-(-1)=m(x-0)
3. All you have left to fill in is m, which is the slope, which is the derivative of the function.
4. Find the derivative to the simplest form. That means get rid of all the x's. In this case, the derivative is 4x. You plug in the x from the given point to get a single number without any variables. x=0, 4*0=0
5. plug in the derivative into m in the point slope form.
6. You now have your answer.

 

The point slope form == the equation of the tangent line.
Same formula.

 

Based on your question, it seems like you're still not comfortable with conceptual notion of derivative. Again, this is why I suggested earlier that you should work through these problems by plotting the functions. It may seem like extra work, but that's how these concepts become internalized.

It may not be feasible to plot everything when dealing with more advanced or abstract math problems. But this is a simple enough example to where actually plotting the graphs and the tangent line is extremely instructive for understanding the concept. Read through my answer to the problem above and then if you still have doubts about what the provided point (0, -1) is used for you can ask.

 

Lol. I never learned it that way in AP Calculus.
Limit Definition is horribly unnecessary.