Out of 20 questions, these were the ones I couldn't really answer
If someone could give me some help with these, that would be great, I don't really understand the whole deal with these questions.
Thanks

Just chain rule

f(x) = 2(x^2-4x)^-2

f'(x) =2 * (-2(x^2-4x)^-3 * (2x-4) ) = -8(x-2)*(x^2 - 4x)^-3, then plug and chug

Could you explain that a little more to me?
I apologize

http://www.youtube.com/watch?v=URHih1NV7rA

or try one of the other many youtube vids

essentially you use u substitution. you replace the complex inside with u and differentiate with respect to u, then multiply by derivative of u with respect to x

so if f(x) = (2x+1)^2, convert this to f(u) = u^2. using the u substitution u = (2x+1). then you differentiate f'(u) = 2u * du/dx

then substitute back your x, remembering to differentiate the inside
u = (2x+1)
du/dx = 2
so f'(x) = 2*(2x+1) * 2

This post was edited by carteblanche on Sep 22 2013 09:49pm

you should re-read your chapter and review Product Rule/Quotient Rule/Chain Rule/etc.

i'll work through #13 for you, its fairly simple

13. In this one, you will use Quotient Rule because its a quotient.

f'(t) = [(t - 6)(4) - (4t + 2)(1)] / (t - 6)^2
f'(t) = [4t - 24 - 4t - 2] / (t - 6)^2
f'(t) = (-26) / (t - 6)^2

now that it's simplified, evaluate the derivative function (what you just found above) at f'(7):

f'(7) = (-26) / (7 - 6)^2
f'(7) = (-26) / (1)^2 = -26

Another possible solution is to plug all of it into Wolfram

so it would be something like -
- 8________
(5-2)^2(5+2)^3

?

Hahaha, how would you plug them in?

Type the expression into the wolfram search bar. Hit enter. It will give you alternate forms, a graph, derivative, integral, and pretty much anything else you could possibly want.

http://www.wolframalpha.com/

what about the (5, 25) ? and the f'(5) part?