Okay, well. I'm pretty sure the second week in Calculus was a trillion times harder than the first. Algebra. I think I need to re-take that class. Or at least a class in patience. Either one will work.
So, before I get myself into a mess, let me clarify something: I'm having a hard time remembering what we did this week and HOW to do it, so if I screw up, I blame my memory.
I think the safest thing for me to explain is the Trig Inverses. I half-way like them. Except the fractions.
Let's try an example problem! Arctan(3x^2+4x)
1. Find and state the formula for Arctan.
1
--------- * u'
1+u^2
2. Replace u for what is inside of the parenthesis.
1
--------------- * (3x^2+4x)'
1+(3x^2+4x)^2
3. Take the derivative of what you are multiplying it by.
(3x^2+4x) = (6x+4)
4. So now you would get
1
--------------- * (6x+4)
1+(3x^2+4x)^2
5. Since (6x+4) = (6x+4)/1, when you multiply, you will get this
(6x+4)
---------------
1+(3x^2+4x)^2
6. Now, foil out (3x^2+4x)^2
(6x+4)
------------------------
1+9x^4 + 24x^3 + 16x^2
7. Simplify what you can and you're finished!
2(3x+2)
------------------------
9x^4 + 24x^3 + 16x^2 + 1
Let's try another half-way simple problem.
Arccos(1/x^2)
1. Find and state the formula for Arccos. (Sense a pattern here?)
-1
----------- * u'
sqrt(1-u^2)
2. Replace u with what is in the parenthesis.
-1
----------------- * (1/x^2)'
sqrt(1-(1/x^2)^2)
3. Take the derivative of what you're multiplying by.
(1/x^2) = (-2/x^3)
4. Multiply to get
2
-------------------
x^3(sqrt(1-(1/x^4))
5. Simplify (completely simplified already) and you're finished!!
I don't know if that was clear or not. I hope so. If someone finds a mistake, let me know please. :)
As for my question...
I completely forgot logs (figures, used to be my favorite).
Can someone explain how to do this problem, I think it's the hardest:
ln ((x^2-1)/x^3)^3
Logs + Fractions + Derivatives /=/ Happiness.
Thanks!!
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let's try this problem
ReplyDeleteso lets take care of the ln first.
so it will be 1/(x^2-1/x^3)^3 now we have to times that by u'.
so this is quotient rule
(x^3) (2x) (2) - (x^2-1) (3x^2) (3x)
-----------------------------------
(x^3)^2
we combine the terms on the top to give us
2 (x^3)(2x) - (3x^3) (x^2-1)
----------------------------
(x^3)^2
now lets combine that with the ln derivative at the top, so it gives us.
3/(x^2-1/x^3)^3 times 2 (x^3)(2x) - (3x^3) (x^2-1)
----------------------------
(x^3)^2
i think this is how you do it.. oh and the 3 on the very top came from the parentheses cubed.. let me know if u see anything wrong
Ok. The only thing I'm seeing wrong with the problem Trina did is that when she combined the u' with the derivative of the ln, she has the (x^3)^2 on the bottom of everything, and if I'm right, I think that should be somewhere else. Oh, and I don't believe that the quotient rule is fully comined So, to completely simplify the quotient rule I got:
ReplyDeletewhen you multiply everything on top you get:
4x^4 - 3x^4 - 3x^3
------------------
x^6
the x^4's cancel, and then you can take an x^3 out of everything. So:
x - 3
-----
x^3
Now, when you multiply this time sthe derivative of ln you should get...
x^7 - 3x^6
----------
(x^2 - 1)^3
Once again, I "THINK" this is right.
ok, so lets take it one step at a time.
ReplyDeletethe first thing you see in that ln problem is the natural log, right? well.. like mrs. robinson said, you work the biggest thing first. so do the natural first, then the second thing you do will be quotient rule.
1. ln(x^2-1/x^3). the formula for natural log is u'/u.
2. u' = [(x^3)(2x)-[(x^2-1)(3x^2)]]/[(x^3)^2]
3. then you would just divide that by u, which is (x^2-1/x^3).
4. you're final problem before simplifying would look like this [[(x^3)(2x)-[(x^2-1)(3x^2)]]/[(x^3)^2]]/[(x^2-1)/(x^3)].
5. you're final step would be to sandwich and simplify.
HOPE THIS HELPED :)