Question

Find the general formula for the derivative.

Answer 1

\[f^{n}(a)\text{ denotes the dervitive at a for the nth derivative} \\find \\f^{27}(1) \text{for} \ f(x) = (1+x^2)^{15}\]

Answer 2

yes
15*14*13*.....
(2x)^n

Answer 3

what happens when your past 15..

Answer 4

wait, isn't there a product rule for 2nd derivative?

Answer 5

yea
the 2

Answer 6

now its even more difficult to generalize...

Answer 7

forgot the exponent 13 on that first part

Answer 8

i don't see a standard or easy way to do this...
http://www.wolframalpha.com/input/?i=27th+derivative+of+%281%2Bx%5E2%29%5E15

Answer 9

waw are you serious.. this was on my exam..

Answer 10

the answers were like 150(28!), 160(28!), (300)(27!)

Answer 11

maybe there's a trick somewhere....lets call for help
@jim_thompson5910 @ganeshie8 @phi @eliassaab

Answer 12

\[f^2 15*14(1+x^{2})^{13}(2x)^{2} + (15)(2)(2x)(1+x^2)^{14} \] I think this is more accurate

Answer 13

I think I would use the binomial theorem to expand (1+x^2)^15 term by term

all the terms with x^n , n<27 will go to zero

Answer 14

hmmm ...

Answer 15

why do my profs want to own me on exams *teary eyes*

Answer 16

\[ (x^2+1)^{15}= \left(\begin{matrix}15 \\ 0\end{matrix}\right)\left( x^2 \right)^{15}1^0+ \left(\begin{matrix}15 \\ 1\end{matrix}\right)\left( x^2 \right)^{14}1^1+ \left(\begin{matrix}15 \\ 2\end{matrix}\right)\left( x^2 \right)^{13}1^2+... \]
\[ x^{30} + 15x^{28} + 105x^{26} + ... \]

Answer 17

well this is definitely new..

Answer 18

If I got the correct pattern for the 27th derivative of x^n
\[ (n (n-1) ... (n-26)) x^{n-27} \]
and
\[ \frac{d^{27}}{dx}x^{30} = (30 \cdot 29\cdot 28... (30-26) )x^{30-27} \\
=(30 \cdot 29\cdot 28... 5\cdot 4) x^3 \\
= \frac{30!}{3!}= \frac{30\cdot 29}{6} \cdot 28!= 145 \cdot 28!
\]

Answer 19

thanks a lot ... now ill just try and figure out what you did..

Answer 20

27th derivative of 15 x^28 is 15 * 28! * x^1 --> 15 * 28!
added together
(145+15) * 28! or

160 * 28!

Answer 21

ofmg !!! I got it right!ahhhhhhhhh

Answer 22

See
http://en.wikipedia.org/wiki/Binomial_theorem#Statement_of_the_theorem
for the binomial theorem

Answer 23

YESSS

Answer 24

!!!

Answer 25

thanks

Answer 26

now I have to learn how to do this as well