Well another week down in calculus and once again I understand some things but I don't understand some things. One thing i understand is get the equation of tangent lines. All you do is take a point and plug it into the derivative of the formula. But if only an x is given then you plug the x into the formula to get your y. And finally you have to put the slope you get when you plug it in the derivative and point in point slope form.
An example is x^3+2x^2-x+1 (2,-1). Then the derivative is 3x^2+4x-1. Now you plug the point in to get the slope. 3(2)^2+4(2)-1. This equals 19 which is your slope. Finally all you do is plug your point and slope into point slope form which is y+1=19(x-2). And even thought this can be condensed into a simpler form if you do it would not be in point slope form. So when you get something like that just leave it.
Also I understand how to get the equation of a normal line. You do the same things as you do for tangent lines except you just make the slope perpendicular. Using the same example above the answer would be y+1=-1/19(x-2). All you do to get the inverse of the slope is take the negative reciprocal.
But on the other hand, I don't understand how to do problems such as ln(lnx^2) or ln(t)/t. I don't even know how to start on these. If someone can get me started and explain it to me it would be greatly appreciated.
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For ln(t)/t
ReplyDeleteThis is a product rule of ln(t) and t^-1
Copy first, derivative of second + Copy second, derivative of first.
ln(t) times (-1/t^2) + (1/t) (1/t) (that's the derivative for ln. 1 over u times u prime. So its just 1/t)
So, simplifying its..
-ln(t) 1
---- + -----
t^2 t^2
so
-ln(t)+1
--------
t^2
is your answer. Hope that helps a bit. :-)
for the ln problems:
ReplyDeletethe fomula is if you have ln (u) then:
1
--- multiplied by "u" prime
(u)
just plug it in..so if you have y=ln(x^2)
i'll be:
1
----- times (2x)
(x^2)
giving you:
2x
-----
(x^2) as your final answer!
hope that was the question and that this helps.
as for the hard ones with all kinds of different things like you having to do the product rule, the "ln" rule, the quotient rule, as well as the exponets kinda freak me out! ha
for the ln problems, the formula is u'/u. or 1/u times u'. just take it one step at a time. if the problem is ln(lnx^2) then it would be the derivative of lnx^2/lnx^2. the derivative of lnx^2 is 2x/x^2. then you would divide that by lnx^2. so you're answer would be ((2x/x^2)/(lnx^2)). then you would just sandwich and simplify! :) HOPE THIS HELPED
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