Mixing
Calculate the $n^{th}$ derivative of this functions?
Given the following function:
$$ f_1(x)=x^3e^x,quadquad f_2(x)=x^2(1+x)^n$$
How to calculate the $n^{th}$ derivative using the General Leibniz rule:
$$ (fg)^{(n)}(x)=sum_{k=0}^nbinom{n}{k}f^{(k)}(x)g^{(n-k)}(x)$$
My work:
let $g(x)=x^3$ and $h(x)=e^x$
and for $kinBbb{N}^*$;
$$ g^{(k)}(x)=
begin{cases}frac{3!}{(3-k)!} x^{3-k}& kleq 3\
0 & k> 3end{cases}$$
and, $$ h^{(k)}(x)=e^x$$
then, $$ f_1^{(n)}(x)=begin{cases}sum_{k=0}^{n}binom{n}{k}frac{3!}{(3-k)!} x^{3-k}e^x & nleq 3 \
sum_{k=0}^{3}binom{3}{k}frac{3!}{(3-k)!} x^{3-k}e^x & n>3
end{cases}$$
The second function i couldn't do it since it involves the power of $n$ and i want to calculate the $n^{th}$ derivative.
Thanks you for your answers and hints.
calculus derivatives
asked Nov 20 '18 at 12:48
hamza boulahia
979319
add a comment |
Given the following function:
$$ f_1(x)=x^3e^x,quadquad f_2(x)=x^2(1+x)^n$$
How to calculate the $n^{th}$ derivative using the General Leibniz rule:
$$ (fg)^{(n)}(x)=sum_{k=0}^nbinom{n}{k}f^{(k)}(x)g^{(n-k)}(x)$$
My work:
let $g(x)=x^3$ and $h(x)=e^x$
and for $kinBbb{N}^*$;
$$ g^{(k)}(x)=
begin{cases}frac{3!}{(3-k)!} x^{3-k}& kleq 3\
0 & k> 3end{cases}$$
and, $$ h^{(k)}(x)=e^x$$
then, $$ f_1^{(n)}(x)=begin{cases}sum_{k=0}^{n}binom{n}{k}frac{3!}{(3-k)!} x^{3-k}e^x & nleq 3 \
sum_{k=0}^{3}binom{3}{k}frac{3!}{(3-k)!} x^{3-k}e^x & n>3
end{cases}$$
The second function i couldn't do it since it involves the power of $n$ and i want to calculate the $n^{th}$ derivative.
Thanks you for your answers and hints.
calculus derivatives
asked Nov 20 '18 at 12:48
hamza boulahia
979319
1
Your result for $f_1$ is not correct, it does not simply become 0. Some of the terms become 0 but that is differemt.
– Ian
Nov 20 '18 at 12:53
For $f_2(x)$, it is probably easiest to expand the function using the binomial theorem. Differentiating a degree $n+2$ polynomial $n$ times should yield a degree $2$ polynomial (with coefficients dependent on $n$) as the answer.
– Kusma
Nov 20 '18 at 12:56
For any $k$, can you compute the $k$th derivative of $f_2$? Do that, then choose $k = n$.
– Mees de Vries
Nov 20 '18 at 13:04
add a comment |
Given the following function:
$$ f_1(x)=x^3e^x,quadquad f_2(x)=x^2(1+x)^n$$
How to calculate the $n^{th}$ derivative using the General Leibniz rule:
$$ (fg)^{(n)}(x)=sum_{k=0}^nbinom{n}{k}f^{(k)}(x)g^{(n-k)}(x)$$
My work:
let $g(x)=x^3$ and $h(x)=e^x$
and for $kinBbb{N}^*$;
$$ g^{(k)}(x)=
begin{cases}frac{3!}{(3-k)!} x^{3-k}& kleq 3\
0 & k> 3end{cases}$$
and, $$ h^{(k)}(x)=e^x$$
then, $$ f_1^{(n)}(x)=begin{cases}sum_{k=0}^{n}binom{n}{k}frac{3!}{(3-k)!} x^{3-k}e^x & nleq 3 \
sum_{k=0}^{3}binom{3}{k}frac{3!}{(3-k)!} x^{3-k}e^x & n>3
end{cases}$$
The second function i couldn't do it since it involves the power of $n$ and i want to calculate the $n^{th}$ derivative.
Thanks you for your answers and hints.
calculus derivatives
asked Nov 20 '18 at 12:48
hamza boulahia
979319
Given the following function:
$$ f_1(x)=x^3e^x,quadquad f_2(x)=x^2(1+x)^n$$
How to calculate the $n^{th}$ derivative using the General Leibniz rule:
$$ (fg)^{(n)}(x)=sum_{k=0}^nbinom{n}{k}f^{(k)}(x)g^{(n-k)}(x)$$
My work:
let $g(x)=x^3$ and $h(x)=e^x$
and for $kinBbb{N}^*$;
$$ g^{(k)}(x)=
begin{cases}frac{3!}{(3-k)!} x^{3-k}& kleq 3\
0 & k> 3end{cases}$$
and, $$ h^{(k)}(x)=e^x$$
then, $$ f_1^{(n)}(x)=begin{cases}sum_{k=0}^{n}binom{n}{k}frac{3!}{(3-k)!} x^{3-k}e^x & nleq 3 \
sum_{k=0}^{3}binom{3}{k}frac{3!}{(3-k)!} x^{3-k}e^x & n>3
end{cases}$$
The second function i couldn't do it since it involves the power of $n$ and i want to calculate the $n^{th}$ derivative.
Thanks you for your answers and hints.
calculus derivatives
calculus derivatives
asked Nov 20 '18 at 12:48
hamza boulahia
979319
asked Nov 20 '18 at 12:48
hamza boulahia
979319
edited Nov 20 '18 at 20:52
asked Nov 20 '18 at 12:48
hamza boulahia
979319
asked Nov 20 '18 at 12:48
hamza boulahia
979319
asked Nov 20 '18 at 12:48
hamza boulahia
979319
979319
1
Your result for $f_1$ is not correct, it does not simply become 0. Some of the terms become 0 but that is differemt.
– Ian
Nov 20 '18 at 12:53
For $f_2(x)$, it is probably easiest to expand the function using the binomial theorem. Differentiating a degree $n+2$ polynomial $n$ times should yield a degree $2$ polynomial (with coefficients dependent on $n$) as the answer.
– Kusma
Nov 20 '18 at 12:56
For any $k$, can you compute the $k$th derivative of $f_2$? Do that, then choose $k = n$.
– Mees de Vries
Nov 20 '18 at 13:04
add a comment |
1
Your result for $f_1$ is not correct, it does not simply become 0. Some of the terms become 0 but that is differemt.
– Ian
Nov 20 '18 at 12:53
For $f_2(x)$, it is probably easiest to expand the function using the binomial theorem. Differentiating a degree $n+2$ polynomial $n$ times should yield a degree $2$ polynomial (with coefficients dependent on $n$) as the answer.
– Kusma
Nov 20 '18 at 12:56
For any $k$, can you compute the $k$th derivative of $f_2$? Do that, then choose $k = n$.
– Mees de Vries
Nov 20 '18 at 13:04
1
1
Your result for $f_1$ is not correct, it does not simply become 0. Some of the terms become 0 but that is differemt.
– Ian
Nov 20 '18 at 12:53
Your result for $f_1$ is not correct, it does not simply become 0. Some of the terms become 0 but that is differemt.
– Ian
Nov 20 '18 at 12:53
For $f_2(x)$, it is probably easiest to expand the function using the binomial theorem. Differentiating a degree $n+2$ polynomial $n$ times should yield a degree $2$ polynomial (with coefficients dependent on $n$) as the answer.
– Kusma
Nov 20 '18 at 12:56
For $f_2(x)$, it is probably easiest to expand the function using the binomial theorem. Differentiating a degree $n+2$ polynomial $n$ times should yield a degree $2$ polynomial (with coefficients dependent on $n$) as the answer.
– Kusma
Nov 20 '18 at 12:56
For any $k$, can you compute the $k$th derivative of $f_2$? Do that, then choose $k = n$.
– Mees de Vries
Nov 20 '18 at 13:04
For any $k$, can you compute the $k$th derivative of $f_2$? Do that, then choose $k = n$.
– Mees de Vries
Nov 20 '18 at 13:04
add a comment |
2 Answers
2
active
oldest
votes
You can only take $2$ derivatives of $x^2$, so you need to take $n-2,n-1,$ or $n$ of the second term. So we have
$$left(x^2(1+x)^nright)^{(n)}=x^2left((1+x)^nright)^{(n)}+2nxleft((1+x)^nright)^{(n-1)}+2{n choose 2}left((1+x)^nright)^{(n-2)}\
=x^2n!+2nxn!(1+x)+2n(n-1)frac {n!}2(1+x)^2$$
answered Nov 20 '18 at 21:30
Ross Millikan
292k23196371
add a comment |
Lets focus on the more general case of $x^alpha e^x$.
Assume that $D^nf(x)=f^{(n)}(x)$.
$$D^0x^alpha=x^alpha$$
$$D^1x^alpha=alpha x^{alpha-1}$$
$$D^2x^alpha=alpha(alpha-1)x^{alpha-2}$$
$$...$$
$$D^kx^alpha=x^{alpha-k}prod_{i=1}^k(alpha-i+1)\=x^{alpha-k}(alpha-1+1)(alpha-2+1)(alpha-3+1)cdots(alpha-n+1)$$
Note that if $alpha$ is an natural number ($1,2,dots$), we have that $k=alpha+1$ is also a natural number, meaning that $D^kx^alpha=0$. It then follows that $D^kx^alpha=0$, provided that $k>alpha$ is a natural number. This is the case because for such values of $k$,
$$prod_{i=1}^k(alpha-i+1)=0$$
My point is, it is sufficient to write
$$D^kx^alpha=x^{alpha-k}prod_{i=1}^k(alpha-i+1)$$
because the $prod_{i=1}^k(alpha-i+1)$ bit automatically encodes the cases of $kleq alpha$, and $k>alpha$. Also, this formula for $D^kx^alpha$ works even if $alpha$ is not a whole number.
You already know that $D^ke^x=e^x$, so we may conclude with
$$D^n(x^alpha e^x)=sum_{k=0}^n{nchoose k}e^xx^{alpha-k}prod_{i=1}^k(alpha-i+1)$$
We can simplify by noting that
$${nchoose k}=prod_{j=1}^kfrac{n-j+1}j$$
Then combining products:
$${nchoose k}prod_{i=1}^k(alpha-i+1)=bigg(prod_{j=1}^kfrac{n-j+1}jbigg)bigg(prod_{i=1}^k(alpha-i+1)bigg)$$
$${nchoose k}prod_{i=1}^k(alpha-i+1)=prod_{i=1}^kfrac{(n-i+1)(alpha-i+1)}{i}$$
$${nchoose k}prod_{i=1}^k(alpha-i+1)=frac1{k!}prod_{i=1}^k(n-i+1)(alpha-i+1)$$
Which gives our simplified result:
$$D^n(x^alpha e^x)=e^xsum_{k=0}^nfrac{x^{alpha-k}}{k!}prod_{i=1}^k(n-i+1)(alpha-i+1)$$
If you are unfamiliar with $prod$ notation,
$$prod_{i=1}^na_i=a_1cdot a_2cdot a_3cdots a_n$$
$$prod_{i=1}^infty a_i=a_1cdot a_2cdot a_3cdots$$
It is exactly the multiplication version of $sum$ notation.
answered Nov 20 '18 at 22:31
clathratus
3,213331
add a comment |
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2 Answers
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You can only take $2$ derivatives of $x^2$, so you need to take $n-2,n-1,$ or $n$ of the second term. So we have
$$left(x^2(1+x)^nright)^{(n)}=x^2left((1+x)^nright)^{(n)}+2nxleft((1+x)^nright)^{(n-1)}+2{n choose 2}left((1+x)^nright)^{(n-2)}\
=x^2n!+2nxn!(1+x)+2n(n-1)frac {n!}2(1+x)^2$$
answered Nov 20 '18 at 21:30
Ross Millikan
292k23196371
add a comment |
You can only take $2$ derivatives of $x^2$, so you need to take $n-2,n-1,$ or $n$ of the second term. So we have
$$left(x^2(1+x)^nright)^{(n)}=x^2left((1+x)^nright)^{(n)}+2nxleft((1+x)^nright)^{(n-1)}+2{n choose 2}left((1+x)^nright)^{(n-2)}\
=x^2n!+2nxn!(1+x)+2n(n-1)frac {n!}2(1+x)^2$$
answered Nov 20 '18 at 21:30
Ross Millikan
292k23196371
add a comment |
You can only take $2$ derivatives of $x^2$, so you need to take $n-2,n-1,$ or $n$ of the second term. So we have
$$left(x^2(1+x)^nright)^{(n)}=x^2left((1+x)^nright)^{(n)}+2nxleft((1+x)^nright)^{(n-1)}+2{n choose 2}left((1+x)^nright)^{(n-2)}\
=x^2n!+2nxn!(1+x)+2n(n-1)frac {n!}2(1+x)^2$$
answered Nov 20 '18 at 21:30
Ross Millikan
292k23196371
You can only take $2$ derivatives of $x^2$, so you need to take $n-2,n-1,$ or $n$ of the second term. So we have
$$left(x^2(1+x)^nright)^{(n)}=x^2left((1+x)^nright)^{(n)}+2nxleft((1+x)^nright)^{(n-1)}+2{n choose 2}left((1+x)^nright)^{(n-2)}\
=x^2n!+2nxn!(1+x)+2n(n-1)frac {n!}2(1+x)^2$$
answered Nov 20 '18 at 21:30
Ross Millikan
292k23196371
answered Nov 20 '18 at 21:30
Ross Millikan
292k23196371
answered Nov 20 '18 at 21:30
Ross Millikan
292k23196371
answered Nov 20 '18 at 21:30
Ross Millikan
292k23196371
292k23196371
add a comment |
add a comment |
Lets focus on the more general case of $x^alpha e^x$.
Assume that $D^nf(x)=f^{(n)}(x)$.
$$D^0x^alpha=x^alpha$$
$$D^1x^alpha=alpha x^{alpha-1}$$
$$D^2x^alpha=alpha(alpha-1)x^{alpha-2}$$
$$...$$
$$D^kx^alpha=x^{alpha-k}prod_{i=1}^k(alpha-i+1)\=x^{alpha-k}(alpha-1+1)(alpha-2+1)(alpha-3+1)cdots(alpha-n+1)$$
Note that if $alpha$ is an natural number ($1,2,dots$), we have that $k=alpha+1$ is also a natural number, meaning that $D^kx^alpha=0$. It then follows that $D^kx^alpha=0$, provided that $k>alpha$ is a natural number. This is the case because for such values of $k$,
$$prod_{i=1}^k(alpha-i+1)=0$$
My point is, it is sufficient to write
$$D^kx^alpha=x^{alpha-k}prod_{i=1}^k(alpha-i+1)$$
because the $prod_{i=1}^k(alpha-i+1)$ bit automatically encodes the cases of $kleq alpha$, and $k>alpha$. Also, this formula for $D^kx^alpha$ works even if $alpha$ is not a whole number.
You already know that $D^ke^x=e^x$, so we may conclude with
$$D^n(x^alpha e^x)=sum_{k=0}^n{nchoose k}e^xx^{alpha-k}prod_{i=1}^k(alpha-i+1)$$
We can simplify by noting that
$${nchoose k}=prod_{j=1}^kfrac{n-j+1}j$$
Then combining products:
$${nchoose k}prod_{i=1}^k(alpha-i+1)=bigg(prod_{j=1}^kfrac{n-j+1}jbigg)bigg(prod_{i=1}^k(alpha-i+1)bigg)$$
$${nchoose k}prod_{i=1}^k(alpha-i+1)=prod_{i=1}^kfrac{(n-i+1)(alpha-i+1)}{i}$$
$${nchoose k}prod_{i=1}^k(alpha-i+1)=frac1{k!}prod_{i=1}^k(n-i+1)(alpha-i+1)$$
Which gives our simplified result:
$$D^n(x^alpha e^x)=e^xsum_{k=0}^nfrac{x^{alpha-k}}{k!}prod_{i=1}^k(n-i+1)(alpha-i+1)$$
If you are unfamiliar with $prod$ notation,
$$prod_{i=1}^na_i=a_1cdot a_2cdot a_3cdots a_n$$
$$prod_{i=1}^infty a_i=a_1cdot a_2cdot a_3cdots$$
It is exactly the multiplication version of $sum$ notation.
answered Nov 20 '18 at 22:31
clathratus
3,213331
add a comment |
Lets focus on the more general case of $x^alpha e^x$.
Assume that $D^nf(x)=f^{(n)}(x)$.
$$D^0x^alpha=x^alpha$$
$$D^1x^alpha=alpha x^{alpha-1}$$
$$D^2x^alpha=alpha(alpha-1)x^{alpha-2}$$
$$...$$
$$D^kx^alpha=x^{alpha-k}prod_{i=1}^k(alpha-i+1)\=x^{alpha-k}(alpha-1+1)(alpha-2+1)(alpha-3+1)cdots(alpha-n+1)$$
Note that if $alpha$ is an natural number ($1,2,dots$), we have that $k=alpha+1$ is also a natural number, meaning that $D^kx^alpha=0$. It then follows that $D^kx^alpha=0$, provided that $k>alpha$ is a natural number. This is the case because for such values of $k$,
$$prod_{i=1}^k(alpha-i+1)=0$$
My point is, it is sufficient to write
$$D^kx^alpha=x^{alpha-k}prod_{i=1}^k(alpha-i+1)$$
because the $prod_{i=1}^k(alpha-i+1)$ bit automatically encodes the cases of $kleq alpha$, and $k>alpha$. Also, this formula for $D^kx^alpha$ works even if $alpha$ is not a whole number.
You already know that $D^ke^x=e^x$, so we may conclude with
$$D^n(x^alpha e^x)=sum_{k=0}^n{nchoose k}e^xx^{alpha-k}prod_{i=1}^k(alpha-i+1)$$
We can simplify by noting that
$${nchoose k}=prod_{j=1}^kfrac{n-j+1}j$$
Then combining products:
$${nchoose k}prod_{i=1}^k(alpha-i+1)=bigg(prod_{j=1}^kfrac{n-j+1}jbigg)bigg(prod_{i=1}^k(alpha-i+1)bigg)$$
$${nchoose k}prod_{i=1}^k(alpha-i+1)=prod_{i=1}^kfrac{(n-i+1)(alpha-i+1)}{i}$$
$${nchoose k}prod_{i=1}^k(alpha-i+1)=frac1{k!}prod_{i=1}^k(n-i+1)(alpha-i+1)$$
Which gives our simplified result:
$$D^n(x^alpha e^x)=e^xsum_{k=0}^nfrac{x^{alpha-k}}{k!}prod_{i=1}^k(n-i+1)(alpha-i+1)$$
If you are unfamiliar with $prod$ notation,
$$prod_{i=1}^na_i=a_1cdot a_2cdot a_3cdots a_n$$
$$prod_{i=1}^infty a_i=a_1cdot a_2cdot a_3cdots$$
It is exactly the multiplication version of $sum$ notation.
answered Nov 20 '18 at 22:31
clathratus
3,213331
add a comment |
Lets focus on the more general case of $x^alpha e^x$.
Assume that $D^nf(x)=f^{(n)}(x)$.
$$D^0x^alpha=x^alpha$$
$$D^1x^alpha=alpha x^{alpha-1}$$
$$D^2x^alpha=alpha(alpha-1)x^{alpha-2}$$
$$...$$
$$D^kx^alpha=x^{alpha-k}prod_{i=1}^k(alpha-i+1)\=x^{alpha-k}(alpha-1+1)(alpha-2+1)(alpha-3+1)cdots(alpha-n+1)$$
Note that if $alpha$ is an natural number ($1,2,dots$), we have that $k=alpha+1$ is also a natural number, meaning that $D^kx^alpha=0$. It then follows that $D^kx^alpha=0$, provided that $k>alpha$ is a natural number. This is the case because for such values of $k$,
$$prod_{i=1}^k(alpha-i+1)=0$$
My point is, it is sufficient to write
$$D^kx^alpha=x^{alpha-k}prod_{i=1}^k(alpha-i+1)$$
because the $prod_{i=1}^k(alpha-i+1)$ bit automatically encodes the cases of $kleq alpha$, and $k>alpha$. Also, this formula for $D^kx^alpha$ works even if $alpha$ is not a whole number.
You already know that $D^ke^x=e^x$, so we may conclude with
$$D^n(x^alpha e^x)=sum_{k=0}^n{nchoose k}e^xx^{alpha-k}prod_{i=1}^k(alpha-i+1)$$
We can simplify by noting that
$${nchoose k}=prod_{j=1}^kfrac{n-j+1}j$$
Then combining products:
$${nchoose k}prod_{i=1}^k(alpha-i+1)=bigg(prod_{j=1}^kfrac{n-j+1}jbigg)bigg(prod_{i=1}^k(alpha-i+1)bigg)$$
$${nchoose k}prod_{i=1}^k(alpha-i+1)=prod_{i=1}^kfrac{(n-i+1)(alpha-i+1)}{i}$$
$${nchoose k}prod_{i=1}^k(alpha-i+1)=frac1{k!}prod_{i=1}^k(n-i+1)(alpha-i+1)$$
Which gives our simplified result:
$$D^n(x^alpha e^x)=e^xsum_{k=0}^nfrac{x^{alpha-k}}{k!}prod_{i=1}^k(n-i+1)(alpha-i+1)$$
If you are unfamiliar with $prod$ notation,
$$prod_{i=1}^na_i=a_1cdot a_2cdot a_3cdots a_n$$
$$prod_{i=1}^infty a_i=a_1cdot a_2cdot a_3cdots$$
It is exactly the multiplication version of $sum$ notation.
answered Nov 20 '18 at 22:31
clathratus
3,213331
Lets focus on the more general case of $x^alpha e^x$.
Assume that $D^nf(x)=f^{(n)}(x)$.
$$D^0x^alpha=x^alpha$$
$$D^1x^alpha=alpha x^{alpha-1}$$
$$D^2x^alpha=alpha(alpha-1)x^{alpha-2}$$
$$...$$
$$D^kx^alpha=x^{alpha-k}prod_{i=1}^k(alpha-i+1)\=x^{alpha-k}(alpha-1+1)(alpha-2+1)(alpha-3+1)cdots(alpha-n+1)$$
Note that if $alpha$ is an natural number ($1,2,dots$), we have that $k=alpha+1$ is also a natural number, meaning that $D^kx^alpha=0$. It then follows that $D^kx^alpha=0$, provided that $k>alpha$ is a natural number. This is the case because for such values of $k$,
$$prod_{i=1}^k(alpha-i+1)=0$$
My point is, it is sufficient to write
$$D^kx^alpha=x^{alpha-k}prod_{i=1}^k(alpha-i+1)$$
because the $prod_{i=1}^k(alpha-i+1)$ bit automatically encodes the cases of $kleq alpha$, and $k>alpha$. Also, this formula for $D^kx^alpha$ works even if $alpha$ is not a whole number.
You already know that $D^ke^x=e^x$, so we may conclude with
$$D^n(x^alpha e^x)=sum_{k=0}^n{nchoose k}e^xx^{alpha-k}prod_{i=1}^k(alpha-i+1)$$
We can simplify by noting that
$${nchoose k}=prod_{j=1}^kfrac{n-j+1}j$$
Then combining products:
$${nchoose k}prod_{i=1}^k(alpha-i+1)=bigg(prod_{j=1}^kfrac{n-j+1}jbigg)bigg(prod_{i=1}^k(alpha-i+1)bigg)$$
$${nchoose k}prod_{i=1}^k(alpha-i+1)=prod_{i=1}^kfrac{(n-i+1)(alpha-i+1)}{i}$$
$${nchoose k}prod_{i=1}^k(alpha-i+1)=frac1{k!}prod_{i=1}^k(n-i+1)(alpha-i+1)$$
Which gives our simplified result:
$$D^n(x^alpha e^x)=e^xsum_{k=0}^nfrac{x^{alpha-k}}{k!}prod_{i=1}^k(n-i+1)(alpha-i+1)$$
If you are unfamiliar with $prod$ notation,
$$prod_{i=1}^na_i=a_1cdot a_2cdot a_3cdots a_n$$
$$prod_{i=1}^infty a_i=a_1cdot a_2cdot a_3cdots$$
It is exactly the multiplication version of $sum$ notation.
answered Nov 20 '18 at 22:31
clathratus
3,213331
answered Nov 20 '18 at 22:31
clathratus
3,213331
answered Nov 20 '18 at 22:31
clathratus
3,213331
answered Nov 20 '18 at 22:31
clathratus
3,213331
3,213331
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1
Your result for $f_1$ is not correct, it does not simply become 0. Some of the terms become 0 but that is differemt.
– Ian
Nov 20 '18 at 12:53
For $f_2(x)$, it is probably easiest to expand the function using the binomial theorem. Differentiating a degree $n+2$ polynomial $n$ times should yield a degree $2$ polynomial (with coefficients dependent on $n$) as the answer.
– Kusma
Nov 20 '18 at 12:56
For any $k$, can you compute the $k$th derivative of $f_2$? Do that, then choose $k = n$.
– Mees de Vries
Nov 20 '18 at 13:04