AP calculas questions AHHH!

[Tyrant222]

Hey, someone please help me. please. im desperate.

I need to learn how to do problems like these for a quiz tomorrow.

Its finding the derivatives.

1) y= csc^2 x + cot^2 x Quotient rule?

2) y=x^3 * sqr(x+1) Product rule?

3) For this one, I have to find the acceleration where x=2. y= (t^3+1)^3
I know I have to take the derivative to get the velocity and then take the derivative again (second derivative) to get the acceleration, then plug in x=2. I can't take the
second derivative without running into problems.

Please some math genius out there help me by showing me the steps.

Edit: plz bump occasionally cause im taking a nap (looong day)
Edit: Plz do not tell me im lazy (I may be so what, less lazy than 300+ students at my hs not taking this class).

I can almost get to the end of the problems. I just dont understand what it takes to finish it.

 

[Ameesh]

look in your book lazy, this is way too easy its just plug and chug

 

[Tyrant222]

plz enlighten me ameesh.

thx

 

[Hakunin]

eh, Level 1 Trig is bad enough for me

 

[yiwonder]

<< 1) y= csc^2 x + cot^2 x Quotient rule? >>


Ok.... Use the power rule on the trig functions, BUT you have to use the chain rule for it to be legal also. See?
2(cscx)(-cscxcotx) + 2(cotx)(-cscx)^2
Clean it up...
-2(cscx)^2cotx - 2(cscx)^2cotx
-4(cscx)^2cotx


<< 2) y=x^3 * sqr(x+1) Product rule? >>


Ok... Use the product rule, but you will have to also use the chain rule.
(x+1)^(1\2)*(3x^2) + x^3*1\2(x+1)^(-1\2)*(1)
See the chain rule? It is the bold...
Just clean it up from there if you need to.


<< y= (t^3+1)^3 >>


The first derivative... Use the power rule and chain rule combo.
y'=3(t^3+1)^2*(3t^2)
Now, you could use the product rule and chain rule again or multiply that out. I'll mulitply it out as it's easier.
y'=9t^8+18t^5+9t^2
Now, take the derivative again...
y''=72t^7+90t^4+18t

Now, just plug 2 into t to find acceleration.
y''=10692

That answer seems pretty odd, but it has to be right.

If you have any other questions or comments, PM me as I'll forget about this thread. My answers have the chance of being wrong, but I do think that they are correct. Good luck. Oh BTW, which AP Calc are you in? AB or BC?

 

[GoldenBear]

1) y= csc^2 x + cot^2 x Quotient rule?

Expanded version:

y' = csc x * csc x + cot x * cot x

You can break it up into two parts, let's deal with the first part first

y1' = csc x (d/dx csc x) + csc x (d/dx csc x)
y1' = csc x (-csc x * cot x) + csc x (-csc x * cot x)
y1' = -2 csc^2 cot x

Then:

y2' = cot x (d/dx cotx) + cot x (d/dx cot x)
y2' = cot x (-csc^2) + cot x (-csc^2)
y2' = -2 cot x csc ^2

Then we have:

(-2 csc^2 cot ) + (-2 cot x csc ^2)

= -4 csc^2 cot x

Simple!

 

[Ender]

E=Mc^2

 

[Tyrant222]

thanks guys, much appreciated

 

[yiwonder]

Ah, Mr. GoldenBear, it isn't tan, it's cot. Check it out if you don't believe me.

 

[20_MuleTeam_Borax]

I don't know if you are required to show your work, but you calculator may be able to calculate (estimate?) derivatives. My Ti-6 can.

 

[GoldenBear]

<< Ah, Mr. GoldenBear, it isn't tan, it's cot. Check it out if you don't believe me. >>

I don't know what you're talking about

 

[yiwonder]

<<

<< Ah, Mr. GoldenBear, it isn't tan, it's cot. Check it out if you don't believe me. >>

I don't know what you're talking about >>


Oh sorry, it must've just been my eyes...

 

[bonkers325]

ahh cal
its bitter

 

[erub]

hehe this is funny, the exact same homework problems I was having problems with tonight also, the Chain Rule

 

[thaRealest]

do u have the 5 step method for derivatives?

 

[Joony]

y=mx+b

...

 

[Daeymion]

<< y=mx+b

... >>



try this instead: Construct a closed set consisting entirely of irrational numbers which is uncountably infinite.

that's a real analysis (calculus) question.