Understanding Derivatives and Tangents: A Beginner's Guide

[scorpa]

Hello everyone,

I have a few questions about derivatives and tangents. Last year I was awesome at derivatives but for some reason it is taking me a bit to bet back into them.

1) If the the tangent line to y=f(x) at (4,3) passes through the point (0,2), find f(4) and f '(4).

For this question I wasn't really sure what to do. But I found the slope of the tangent using the slope forumula and the points (4,3) and (0,2). I found this slope to be (1/4). Then I found the equation of the tangent line to be y=(1/4)x+2. I'm not really sure where to go with this now. If the equation I just figured out is the equation of the tangent line it is the equation of the derivative of the function so to find f '(4) I should just be able to stick 4 into that right? But how would I find f(4)? I must be doing something wrong because I know I shouldn't have to get into intergrals yet...that doesn't come for awhile. I think I must be misunderstanding the question or something.

2) The limit represents the derivative of some function f at some number a. State such an f and a this case. (lim as x approaches pi/4) (tanx-1)/(x-pi/4).
I have absolutely no idea how to go about this one, limits and derivatives of trig functions has always been a weak spot for me.

There is one more question that I have but I am going to try and keep working on it before I ask it here, I think I might be close to understanding it.
Thanks a lot guys.

 

[TD]

If (4,3) is a point of the function y=f(x), then surely f(4) will be 3, no?
For f'(4), you need the slope of the tangent line there, which you calculated correctly I believe.

The limit for the definition you'll be using here is the following one:

$$f'\left( a \right) = \mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - f\left( a \right)}}<br /> {{x - a}}$$

Can you now identify f and a?

 

[scorpa]

Oh ok, I get that first part now, thanks a lot. What about the second one?

 

[scorpa]

Ok so I know for the second one that you have to break the tan up into sin/cos-1 and then get the one over a common factor so that you end up having:
(sinx-cosx/cosx)x(1/(x-pi/4)) but after that I get lost and do not know where to go. Any suggestions?
-Thanks

 

[krab]

I suggest you read TD's post again, and answer his question. You don't have to use tan= sin/cos. Just compare your limit equation with TD's and from this you will know exactly what f and a are.

 

[scorpa]

would f(x) be the entire equation and f(a) be pi/4? Is that what you mean?

 

[scorpa]

Oh and I have one more question for the top part, once I find the equation of the tangent line do I have to differentiate that to find f'(4) ?

 

[scorpa]

Ahhh for the second question they only want you to state what f and a are not what the derivative of the thing is...good lord I can be stupid sometimes. Anyway I am still stuck the first one, do I have to do what I said above?

 

[scorpa]

bump

Ok, I ended up getting that f(4) and f'(4) both equal 3. I guess this makes sense, seeing as the tangent line and actual curve are both at the same point at the point where the tangent hits the curve. So this should be right?

 

[TD]

would f(x) be the entire equation and f(a) be pi/4? Is that what you mean?

Compare

$$\mathop {\lim }\limits_{x \to \pi /4} \frac{{\tan x - 1}}<br /> {{x - \pi /4}}$$

with

$$\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - f\left( a \right)}}<br /> {{x - a}}$$

 

[HallsofIvy]

bump

Ok, I ended up getting that f(4) and f'(4) both equal 3.

Is this about problem 1 again? I thought you were already told how to do that problem.
I thought you had already calculated that the slope of the tangent line was
$$\frac{3- 2}{4- 0}= \frac{1}{4}$$

I guess this makes sense, seeing as the tangent line and actual curve are both at the same point at the point where the tangent hits the curve. So this should be right?

 

[NascentOxygen]

HallsofIvy breathing new life into a thread whose author believed it to be dead and buried almost 10 years ago ...

 

[HallsofIvy]

Its never too late!