Question

(b) If F(x)=f(x)g(x) find F'''(x) and F^(4)(x)
(c) Guess formula for F^(n)

Answer 1

@zepdrix

Answer 2

i know F'(x)=fg'+gf' and F"(x)=f"g+2f'g'+fg"

Answer 3

Do you see a pattern forming?

Answer 4

binomial expansion?

Answer 5

That's what I'd say, but just to make sure: the third derivative would be
\[F'''(x)=f'''g+f''g'+2f''g'+2f'g''+f'g''+fg'''\\
F'''(x)=f'''g+3f''g'+3f'g''+fg'''\]
Yep, I'd say it's a safe bet.

Answer 6

so how would i write out the binomial expansion

Answer 7

\[F^{(3)}=\begin{pmatrix}3\\0\end{pmatrix}f^{(3)}g^{(0)}+\begin{pmatrix}3\\1\end{pmatrix}f^{(2)}g^{(1)}+\begin{pmatrix}3\\2\end{pmatrix}f^{(1)}g^{(2)}+\begin{pmatrix}3\\3\end{pmatrix}f^{(0)}g^{(3)}\]
Try forming an \(n\)-th degree pattern from that.

Answer 8

... where \(g^{(0)}\) is the "zeroeth" derivative, meaning the function itself.

Answer 9

\[F ^{n}=\left(\begin{matrix}n \\ 0\end{matrix}\right)f ^{n-1}...\]?

Answer 10

Maybe if you start off by writing as a summation:

\(\color{blue}{\text{Originally Posted by}}\) @SithsAndGiggles
\[F^{(3)}=\sum_{k=0}^3\begin{pmatrix}3\\k\end{pmatrix}f^{(3-k)}g^{(k)}\]
Try forming an \(n\)-th degree pattern from that.
\(\color{blue}{\text{End of Quote}}\)

Answer 11

Why not?\[
\sum_{k=0}^n\binom nkf^{(k)}g^{(n-k)}
\]

Answer 12

ok thanks. but what would the fourth derivative be

Answer 13

Well, you can either continue to differentiate from \(F^{(3)}\), or you can apply the formula you've found:
\[F^{(4)}=\sum_{k=0}^4\binom 4kf^{(4-k)}g^{(4)}=\sum_{k=0}^4\binom 4kf^{(k)}g^{(4-k)}\]

Answer 14

but i would need to expand that. what would k be?

Answer 15

k = 0, 1, 2, 3, 4.

Answer 16

the first k in \[\left(\begin{matrix}4 \\ k\end{matrix}\right)\]

Answer 17

Is this your first time working with indices of a sum? Refer to @ranga's post. Plug in each \(k\). You'll have a total of 5 terms.

Answer 18

but what would the first k be

Answer 19

\[F ^{(4)} = \left(\begin{matrix}4 \\ 0\end{matrix}\right)f ^{(4)}g ^{(0)} + \left(\begin{matrix}4 \\ 1\end{matrix}\right)f ^{(3)}g ^{(1)} + ......\]

I have expanded the first two terms. There are three more to go.

Answer 20

sorry but i dont remember binomial thereom. please help me out

Answer 21

You have a general formula for the nth derivative given by SithsAndGiggles and wio above.
It is a sigma meaning summation.
For the fourth derivative n = 4.
First put k = 0 and you get the first term as I did above.
Then put k = 1 and you will get the second term also shown above.
Then put k = 2 and you will get the third term
and then k = 3 and k = 4 to get the fourth and fifth term.
Add them all up (sigma) and you will get the fourth derivative.

Answer 22

what do you do with the \[\left(\begin{matrix}4 \\ 1\end{matrix}\right)\] or with the \[\left(\begin{matrix}4 \\ 0\end{matrix}\right)\]

Answer 23

@ranga i dont know what to do with that

Answer 24

\[\left(\begin{matrix}n \\ r\end{matrix}\right) = \frac{ n! }{ r!(n-r)!}\]

Answer 25

?

Answer 26

(n factorial) / (r factorial times (n-r) factorial)

5! = 5 x 4 x 3 x 2 x 1 = 120
4! = 4 x 3 x 2 x 1 = 24
0! = 1

Answer 27

what would \[\left(\begin{matrix}4 \\ 0\end{matrix}\right)\] be

Answer 28

4! / (0! x 4!) = 1

Answer 29

ok thanks

Answer 30

you are welcome.