taylor series/n-th derivative ahh!

[voidflux]

Hello everyone!
i'm having probems figuring these 2 problems out: The first one is a taylor series and hte 2nd one is wanting me to find the nth-derivative of f when, f(x) = 4*ln(4*x+2);

It says, first find the A and so that

21+9x             A                B
--------   = -----------   +  -----------
7 + x          7+x              7+x

so i found the A = 2, and B = 9,
So i have

21+9x
--------
7+x

and it says, then use this to determine the Taylor Series for
21+9x
-------
7+x


the options i have are:
E stands for sumnation

A. (3-6)*[E k=1 to infinity (-1)^k-1 (x^k)/(7^k)]
B. (3+6)*[E k=1 to infinity (-1)^k-1 (x^k)(7^k)]
C. 6*[E k = 1 to infinity (-1)^k-1 (x^k)/(7^k)]
D. (3+6)*[E k=1 to infinity (-1)^k-1 (x^k)/(7^k)]
E. (3+6)*[E k = 1 to infinity (x^k)/(7^k)]

and the 2nd one:
find the nth-derivative of f when f(x) = 4*ln(4*x +2);
What am i suppose to do here exactly? I have maple and I tried to do a taylor series problem and it looks nothing like the choices i have.

Thanks!

I was taught basic taylor series, this is why i have no idea on what they even want, I was never shown how to figure out the series itself, just evaluating it

[jemidiah]

For the second one:

4*ln(u) 'For simplicity, u = 4x + 2

1st derivative:
4*u'/u = 16*u^-1

2nd derivative:
Power rule: (axⁿ)' = anx^n-1 so
-16*u^-2

3rd derivative:
32*u^-3

So:

nth derivative:
16*((n-1)!)*(-1)ⁿ⁺¹/uⁿ

Resubstituting 4x + 2 for u:
16*((n-1)!)*(-1)ⁿ⁺¹/(4x+2)ⁿ

16(n-1)! * -1ⁿ⁺¹
----------------
(4x+2)ⁿ

[voidflux]

Hey thanks for the reply, my choices for the 2nd one are:

A.  (n-1)! (4)(4^n)
        ---------------
           (4x+2)^n

B.  (-1)^(n-1)  (n!)(4)(4^n)
        --------------------------
           (4x+2)^n

C.  (-1)^(n-1)  (n-1!)(4)(4^n)
        -----------------------------
           (4x+2)^n

D.  (-1)^(n-1)  (n-1!)(4)
        -----------------------------
           (4x+2)^n

I used ur method and i got the same answer as you but i don't see it listed,

[jemidiah]

Whoops; I forgot about chain-ruling it with that constant;

4*ln(4x+2)

1st derivative:
4*4/(4x+2) = 16*(4x+2)^-1

2nd derivative:
Power rule: (auⁿ)' = anu'u^n-1 so
4*4*-1*4*u^-2 = -64*u^-2

3rd derivative:
4*4*4*-1*-2*u^-3 = 128*u^-3

So:

nth derivative:
4*(n-1)!*(-1)ⁿ⁺¹*(u')ⁿ/uⁿ

Resubstituting 4x + 2 for u:
4*(n-1)!*(-1)ⁿ⁺¹*4ⁿ/(4x+2)ⁿ


So all that for C.


Note: I edited this late at night, so, while the conclusion is correct, the actual steps might have some flaws.

[voidflux]

Thanks alot!