Ok, so we go back to school soon, which is depressing. Now instead of just doing blogs, we actually have to learn stuff and have homework again :/
Today I'll talk about a definition of a derivative, product rule, quotient rule, and how to find tangent lines.
Instantaneous speed, or a definition of a derivative is described with this formula:
limn-0 f(x+h)-f(x)/h
As I said before, you can use this to find instantaneous speed. For example:
If you are asked to find instantaneous speed at x=2 and y=2x^2, you just plug it in to the formula of the definition of a derivative.
limn-0 16(2+h)^2 - 16(2)^2/h
= 16(4 + 4h + h^2) - 64
= 64 + 64h + 16h^2 - 64
= 64 u/sec
Product rule is just another way to find a derivative when something is multiplied together. The formula for this is uv' + vu'. This means to coppy the first term, times the derivative of the second term, plus coppy the second term, times the derivative of the first term. Other than taking a regular derivative with nothing multiplied or devided by it, product rule is pretty easy if the problem isn't huge.
For example:
(x^2)(3x)
= x^2(3) + 3x(2x)
= 3x^2 + 6x^2
= 9x^2
Quotient rule is a way of taking a derivative when it is in a fraction and it cannot be broken up or simplified. The formula for this is uv' - vu'/v^2. This means to coppy the first term down, timesed by the derivative of the second term, minus coppy the second term down, timesed by the derivative of the first term, all over the second term squared. The quotient rule is also pretty easy if the problem isn't huge. It can get confusing with many functions in it. You should not use quotient rule if other things can be done to the derivative to get around it. You should not use a quotient rule if you have a problem such as 5/x^2
For tangent lines, you plug information given by the problem into the point slope formula. You plug the exact numbers in for x and y. To find m, which is your slope, you will need to take the derivative of the two (a derivative is a slope). I used to have problems with tangent line problems. I don't remember why, because they are sooo easy.
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If you didn't quite understand mamies description heres an example
ReplyDeleteInstantaneous speed:
lim f(x+h)-f(x)
---------------
h->0 h
Find the instantaneous speed at t=2
y=16t^2 lim 16(2+h)^2-16(2)^2
f(t)=16t^2 ------------------------ =64
h->0 h
Wow...your description of the tangent lines really helped! I was still having trouble with them! That's got to be the simplest explanation ever! Thanks!! *applaud*
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