Find the slope of the function at the given value: y=sin(2x) atx=−π
Find the derivative of the given function. Plug in x = -π
x = -pi .-.
I did the problem, but I am not sure where I went wrong
And we haven't gotten to derivatives yet
We are on slopes of tangent and secant lines and stuff like that. I understand it is similar to derivatives, but we haven't actually gotten there yet. We need to use the average and instantaneous rates of change stuff
This is asking for the slope at a specific point, which is instantaneous rate. Average rate would be the average rate of change between two points. I'm not sure how to calculate instantaneous rate at a point without the derivative, I think.
Right, so thats the tangent line
Ohh, yes, yes. The slope of the line tangent to the function at x = -pi
yes
Its a kutasoftware problem : http://cdn.kutasoftware.com/Worksheets/Calc/04%20-%20Slope%20at%20a%20Value.pdf
Bleh. I wonder if there is a process I'm forgetting about.
It involves
|dw:1410134769258:dw|
The slope of the tangent line is equivalent to the value of the derivative at the same point. Oh, ok, good. That's the limit definition of the derivative.
Ok
If you follow the definition of the derivative to your given function, applying the limit and such, you'll get the derivative. Then, you can plug in -pi for x.
I did. Idk where I went wrong. I got 0 instead of 2. I was hoping someone could show me how to do it
@girlover
Ok, so my problem comes down to lim (h->0) sin(2x)/x
How does that equal 2?
Can someone show me how they solved it? I keep messing up
@jim_thompson5910
Please, I have had this question up for like 2 hours
@Hero
@satellite73
@Ashleyisakitty
I'm assuming you mean at x=\(-\pi\)
yes
I accidentally left out the space
No problem, let me work this out right quick.
Thanks so much!
Im going to go to bed soon, so please ASAP
\(\frac{ð (y = sin(2 x))}{ð x} ~where~ x = -\pi\) plug in \(\frac{ð (y = sin(2 \times-\pi))}{ð -\pi}\) \(y'(-\pi) = 2\)
It is blurring out the symbols
Refresh the page
ok
i dont understand what you did. we have to use the instantaneous rate of change formula
That was the best I could find as an explanation, I know it's no help; it's better than nothing, I suppose.
I appreciate it
@jim_thompson5910
lol 40 minutes until i can bump
I have to go to bed, i will respond in the morning.
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