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Finding the $n$-th derivative of $a_nx^n+a_n-1x^n-1+cdots+a_1x+a_0$
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Let $f(x)= a_nx^n+a_n-1x^n-1+cdots+a_1x+a_0$.
I am trying to find $f^n(x)$.
By applying the power rule $n$ times, I get this
$$f^n(x)=a_n(ncdot n)x^n-n+cdots+ a_1$$
which I think can be simplified to
$$f^n(x)=a_nn^2+cdots+ a_1$$
However, I don't think I have the correct answer, as my exercise book is telling me the answer is.
$$a_n,ncdot(n-1)cdot,cdots,cdot 1$$
What did I do wrong? I'm under the impression I have done multiple mistakes.
derivatives
edited Jul 30 at 0:43
Blue
43.6k868141
asked Jul 30 at 0:32
Cedric Martens
24329
add a comment |Â
up vote
3
down vote
favorite
Let $f(x)= a_nx^n+a_n-1x^n-1+cdots+a_1x+a_0$.
I am trying to find $f^n(x)$.
By applying the power rule $n$ times, I get this
$$f^n(x)=a_n(ncdot n)x^n-n+cdots+ a_1$$
which I think can be simplified to
$$f^n(x)=a_nn^2+cdots+ a_1$$
However, I don't think I have the correct answer, as my exercise book is telling me the answer is.
$$a_n,ncdot(n-1)cdot,cdots,cdot 1$$
What did I do wrong? I'm under the impression I have done multiple mistakes.
derivatives
edited Jul 30 at 0:43
Blue
43.6k868141
asked Jul 30 at 0:32
Cedric Martens
24329
1
Start with $,left(x^nright)' = n x^n-1,$, $,left(x^nright)'' = n(n-1) x^n-2,$, $,left(x^nright)''' = n(n-1)(n-2) x^n-3 ,ldots,$
– dxiv
Jul 30 at 0:34
Hint: $(x^n)' = n x^n-1$, but $(nx^n-1)' = n(n-1) x^n-2$, and so on. Repeated application of the power rule give a falling factorial coefficient.
– Xander Henderson
Jul 30 at 0:35
The exercise book should say the answer is $a_nn(n-1)...1$.
– é«˜ç”°èˆª
Jul 30 at 0:39
You are right, that is the answer in the book. I badly transcribed it, I modified my question. Thank you
– Cedric Martens
Jul 30 at 0:41
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $f(x)= a_nx^n+a_n-1x^n-1+cdots+a_1x+a_0$.
I am trying to find $f^n(x)$.
By applying the power rule $n$ times, I get this
$$f^n(x)=a_n(ncdot n)x^n-n+cdots+ a_1$$
which I think can be simplified to
$$f^n(x)=a_nn^2+cdots+ a_1$$
However, I don't think I have the correct answer, as my exercise book is telling me the answer is.
$$a_n,ncdot(n-1)cdot,cdots,cdot 1$$
What did I do wrong? I'm under the impression I have done multiple mistakes.
derivatives
edited Jul 30 at 0:43
Blue
43.6k868141
asked Jul 30 at 0:32
Cedric Martens
24329
Let $f(x)= a_nx^n+a_n-1x^n-1+cdots+a_1x+a_0$.
I am trying to find $f^n(x)$.
By applying the power rule $n$ times, I get this
$$f^n(x)=a_n(ncdot n)x^n-n+cdots+ a_1$$
which I think can be simplified to
$$f^n(x)=a_nn^2+cdots+ a_1$$
However, I don't think I have the correct answer, as my exercise book is telling me the answer is.
$$a_n,ncdot(n-1)cdot,cdots,cdot 1$$
What did I do wrong? I'm under the impression I have done multiple mistakes.
derivatives
edited Jul 30 at 0:43
Blue
43.6k868141
asked Jul 30 at 0:32
Cedric Martens
24329
edited Jul 30 at 0:43
Blue
43.6k868141
edited Jul 30 at 0:43
Blue
43.6k868141
edited Jul 30 at 0:43
Blue
43.6k868141
43.6k868141
asked Jul 30 at 0:32
Cedric Martens
24329
asked Jul 30 at 0:32
Cedric Martens
24329
asked Jul 30 at 0:32
Cedric Martens
24329
24329
1
Start with $,left(x^nright)' = n x^n-1,$, $,left(x^nright)'' = n(n-1) x^n-2,$, $,left(x^nright)''' = n(n-1)(n-2) x^n-3 ,ldots,$
– dxiv
Jul 30 at 0:34
Hint: $(x^n)' = n x^n-1$, but $(nx^n-1)' = n(n-1) x^n-2$, and so on. Repeated application of the power rule give a falling factorial coefficient.
– Xander Henderson
Jul 30 at 0:35
The exercise book should say the answer is $a_nn(n-1)...1$.
– é«˜ç”°èˆª
Jul 30 at 0:39
You are right, that is the answer in the book. I badly transcribed it, I modified my question. Thank you
– Cedric Martens
Jul 30 at 0:41
add a comment |Â
1
Start with $,left(x^nright)' = n x^n-1,$, $,left(x^nright)'' = n(n-1) x^n-2,$, $,left(x^nright)''' = n(n-1)(n-2) x^n-3 ,ldots,$
– dxiv
Jul 30 at 0:34
Hint: $(x^n)' = n x^n-1$, but $(nx^n-1)' = n(n-1) x^n-2$, and so on. Repeated application of the power rule give a falling factorial coefficient.
– Xander Henderson
Jul 30 at 0:35
The exercise book should say the answer is $a_nn(n-1)...1$.
– é«˜ç”°èˆª
Jul 30 at 0:39
You are right, that is the answer in the book. I badly transcribed it, I modified my question. Thank you
– Cedric Martens
Jul 30 at 0:41
1
1
Start with $,left(x^nright)' = n x^n-1,$, $,left(x^nright)'' = n(n-1) x^n-2,$, $,left(x^nright)''' = n(n-1)(n-2) x^n-3 ,ldots,$
– dxiv
Jul 30 at 0:34
Start with $,left(x^nright)' = n x^n-1,$, $,left(x^nright)'' = n(n-1) x^n-2,$, $,left(x^nright)''' = n(n-1)(n-2) x^n-3 ,ldots,$
– dxiv
Jul 30 at 0:34
Hint: $(x^n)' = n x^n-1$, but $(nx^n-1)' = n(n-1) x^n-2$, and so on. Repeated application of the power rule give a falling factorial coefficient.
– Xander Henderson
Jul 30 at 0:35
Hint: $(x^n)' = n x^n-1$, but $(nx^n-1)' = n(n-1) x^n-2$, and so on. Repeated application of the power rule give a falling factorial coefficient.
– Xander Henderson
Jul 30 at 0:35
The exercise book should say the answer is $a_nn(n-1)...1$.
– é«˜ç”°èˆª
Jul 30 at 0:39
The exercise book should say the answer is $a_nn(n-1)...1$.
– é«˜ç”°èˆª
Jul 30 at 0:39
You are right, that is the answer in the book. I badly transcribed it, I modified my question. Thank you
– Cedric Martens
Jul 30 at 0:41
You are right, that is the answer in the book. I badly transcribed it, I modified my question. Thank you
– Cedric Martens
Jul 30 at 0:41
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Recall that the power rule for differentiation states:
Let $f(x)=x^n$ for some $ninmathbbRbackslashlbrace 0rbrace$. Then,
$$fracddxf(x)=nx^n-1$$
Also recall that differentiation is linear in the sense that $$fracddx[f_1(x)+f_2(x)+cdots+f_k(x)]=fracddxf_1(x)+fracddxf_2(x)+cdots+fracddxf_k(x)$$
Now consider your function $f(x)=a_nx^n+a_n-1x^n-1+cdots+a_1x+a_0$. Differentiating and applying linearity, we obtain
$$fracddxf(x)=fracddxa_nx^n+fracddxa_n-1x^n-1+cdots+fracddxa_1x+fracddxa_0$$
The constant term at the end becomes $0$, since the derivative of a constant is always $0$. The second to last term, $fracddxa_1x$ is simply $a_1$, since the derivative of a line is always its slope. As for the other terms, we simply apply the power rule. After completing the first round of differentiation, we should have something like this:
$$f^1(x)=a_nnx^n-1+a_n-1x^n-2+cdots+a_2x+a_1$$
If we keep repeating the differentiation process, you will see that there is always a constant term $a_m$ at the end which just becomes $0$. Since $f(x)$ has $n+1$ terms (each $a_ix^i$ from $i=0$ up to $i=n$), after $n$ rounds of differentiation, you should have eliminated all the terms except for the very first one: $a_nx^n$. Applying the power rule to just that term, we get:
beginalign
fracddxa_nx_n &=a_nnx^n-1\
fracddxa_nnx^n-1 & = a_nn(n-1)x^n-2\
fracddxa_nn(n-1)x^n-2& =a_nn(n-1)(n-2)x^n-3
\ &vdots
\ fracddxa_nn(n-1)(n-2)cdots(3)x^2& =a_nn(n-1)(n-2)cdots(3)(2)x
\ fracddxa_nn(n-1)(n-2)cdots(3)(2)x&=a_nn(n-1)(n-2)cdots(3)(2)(1)
\ & = a_ncdot n!
endalign
answered Jul 30 at 1:10
高田航
1,116318
add a comment |Â
up vote
0
down vote
First, you should convince yourself that all the terms on the right after the first do not contribute because you eventually take the derivative of a constant, which is zero.
Then do it by hand for $n=3$. We want the third derivative of $a_3x^3$. The first derivative is $3a_3x^2$, the second is $3cdot 2 a_3x$, the third is $3cdot 2 cdot 1a_3=3!a_3$
Now convince yourself that this is general and $f^n(x)=a_nn!$
Your proposed answer has two problems: the exponent decreases by $1$ each time you take a derivative so the second one should be $n-1$ and there are $n$ derivatives, not $2$. The numbers you bring down are multiplied, not added.
answered Jul 30 at 1:00
Ross Millikan
275k21185351
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Recall that the power rule for differentiation states:
Let $f(x)=x^n$ for some $ninmathbbRbackslashlbrace 0rbrace$. Then,
$$fracddxf(x)=nx^n-1$$
Also recall that differentiation is linear in the sense that $$fracddx[f_1(x)+f_2(x)+cdots+f_k(x)]=fracddxf_1(x)+fracddxf_2(x)+cdots+fracddxf_k(x)$$
Now consider your function $f(x)=a_nx^n+a_n-1x^n-1+cdots+a_1x+a_0$. Differentiating and applying linearity, we obtain
$$fracddxf(x)=fracddxa_nx^n+fracddxa_n-1x^n-1+cdots+fracddxa_1x+fracddxa_0$$
The constant term at the end becomes $0$, since the derivative of a constant is always $0$. The second to last term, $fracddxa_1x$ is simply $a_1$, since the derivative of a line is always its slope. As for the other terms, we simply apply the power rule. After completing the first round of differentiation, we should have something like this:
$$f^1(x)=a_nnx^n-1+a_n-1x^n-2+cdots+a_2x+a_1$$
If we keep repeating the differentiation process, you will see that there is always a constant term $a_m$ at the end which just becomes $0$. Since $f(x)$ has $n+1$ terms (each $a_ix^i$ from $i=0$ up to $i=n$), after $n$ rounds of differentiation, you should have eliminated all the terms except for the very first one: $a_nx^n$. Applying the power rule to just that term, we get:
beginalign
fracddxa_nx_n &=a_nnx^n-1\
fracddxa_nnx^n-1 & = a_nn(n-1)x^n-2\
fracddxa_nn(n-1)x^n-2& =a_nn(n-1)(n-2)x^n-3
\ &vdots
\ fracddxa_nn(n-1)(n-2)cdots(3)x^2& =a_nn(n-1)(n-2)cdots(3)(2)x
\ fracddxa_nn(n-1)(n-2)cdots(3)(2)x&=a_nn(n-1)(n-2)cdots(3)(2)(1)
\ & = a_ncdot n!
endalign
answered Jul 30 at 1:10
高田航
1,116318
add a comment |Â
up vote
2
down vote
accepted
Recall that the power rule for differentiation states:
Let $f(x)=x^n$ for some $ninmathbbRbackslashlbrace 0rbrace$. Then,
$$fracddxf(x)=nx^n-1$$
Also recall that differentiation is linear in the sense that $$fracddx[f_1(x)+f_2(x)+cdots+f_k(x)]=fracddxf_1(x)+fracddxf_2(x)+cdots+fracddxf_k(x)$$
Now consider your function $f(x)=a_nx^n+a_n-1x^n-1+cdots+a_1x+a_0$. Differentiating and applying linearity, we obtain
$$fracddxf(x)=fracddxa_nx^n+fracddxa_n-1x^n-1+cdots+fracddxa_1x+fracddxa_0$$
The constant term at the end becomes $0$, since the derivative of a constant is always $0$. The second to last term, $fracddxa_1x$ is simply $a_1$, since the derivative of a line is always its slope. As for the other terms, we simply apply the power rule. After completing the first round of differentiation, we should have something like this:
$$f^1(x)=a_nnx^n-1+a_n-1x^n-2+cdots+a_2x+a_1$$
If we keep repeating the differentiation process, you will see that there is always a constant term $a_m$ at the end which just becomes $0$. Since $f(x)$ has $n+1$ terms (each $a_ix^i$ from $i=0$ up to $i=n$), after $n$ rounds of differentiation, you should have eliminated all the terms except for the very first one: $a_nx^n$. Applying the power rule to just that term, we get:
beginalign
fracddxa_nx_n &=a_nnx^n-1\
fracddxa_nnx^n-1 & = a_nn(n-1)x^n-2\
fracddxa_nn(n-1)x^n-2& =a_nn(n-1)(n-2)x^n-3
\ &vdots
\ fracddxa_nn(n-1)(n-2)cdots(3)x^2& =a_nn(n-1)(n-2)cdots(3)(2)x
\ fracddxa_nn(n-1)(n-2)cdots(3)(2)x&=a_nn(n-1)(n-2)cdots(3)(2)(1)
\ & = a_ncdot n!
endalign
answered Jul 30 at 1:10
高田航
1,116318
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Recall that the power rule for differentiation states:
Let $f(x)=x^n$ for some $ninmathbbRbackslashlbrace 0rbrace$. Then,
$$fracddxf(x)=nx^n-1$$
Also recall that differentiation is linear in the sense that $$fracddx[f_1(x)+f_2(x)+cdots+f_k(x)]=fracddxf_1(x)+fracddxf_2(x)+cdots+fracddxf_k(x)$$
Now consider your function $f(x)=a_nx^n+a_n-1x^n-1+cdots+a_1x+a_0$. Differentiating and applying linearity, we obtain
$$fracddxf(x)=fracddxa_nx^n+fracddxa_n-1x^n-1+cdots+fracddxa_1x+fracddxa_0$$
The constant term at the end becomes $0$, since the derivative of a constant is always $0$. The second to last term, $fracddxa_1x$ is simply $a_1$, since the derivative of a line is always its slope. As for the other terms, we simply apply the power rule. After completing the first round of differentiation, we should have something like this:
$$f^1(x)=a_nnx^n-1+a_n-1x^n-2+cdots+a_2x+a_1$$
If we keep repeating the differentiation process, you will see that there is always a constant term $a_m$ at the end which just becomes $0$. Since $f(x)$ has $n+1$ terms (each $a_ix^i$ from $i=0$ up to $i=n$), after $n$ rounds of differentiation, you should have eliminated all the terms except for the very first one: $a_nx^n$. Applying the power rule to just that term, we get:
beginalign
fracddxa_nx_n &=a_nnx^n-1\
fracddxa_nnx^n-1 & = a_nn(n-1)x^n-2\
fracddxa_nn(n-1)x^n-2& =a_nn(n-1)(n-2)x^n-3
\ &vdots
\ fracddxa_nn(n-1)(n-2)cdots(3)x^2& =a_nn(n-1)(n-2)cdots(3)(2)x
\ fracddxa_nn(n-1)(n-2)cdots(3)(2)x&=a_nn(n-1)(n-2)cdots(3)(2)(1)
\ & = a_ncdot n!
endalign
answered Jul 30 at 1:10
高田航
1,116318
Recall that the power rule for differentiation states:
Let $f(x)=x^n$ for some $ninmathbbRbackslashlbrace 0rbrace$. Then,
$$fracddxf(x)=nx^n-1$$
Also recall that differentiation is linear in the sense that $$fracddx[f_1(x)+f_2(x)+cdots+f_k(x)]=fracddxf_1(x)+fracddxf_2(x)+cdots+fracddxf_k(x)$$
Now consider your function $f(x)=a_nx^n+a_n-1x^n-1+cdots+a_1x+a_0$. Differentiating and applying linearity, we obtain
$$fracddxf(x)=fracddxa_nx^n+fracddxa_n-1x^n-1+cdots+fracddxa_1x+fracddxa_0$$
The constant term at the end becomes $0$, since the derivative of a constant is always $0$. The second to last term, $fracddxa_1x$ is simply $a_1$, since the derivative of a line is always its slope. As for the other terms, we simply apply the power rule. After completing the first round of differentiation, we should have something like this:
$$f^1(x)=a_nnx^n-1+a_n-1x^n-2+cdots+a_2x+a_1$$
If we keep repeating the differentiation process, you will see that there is always a constant term $a_m$ at the end which just becomes $0$. Since $f(x)$ has $n+1$ terms (each $a_ix^i$ from $i=0$ up to $i=n$), after $n$ rounds of differentiation, you should have eliminated all the terms except for the very first one: $a_nx^n$. Applying the power rule to just that term, we get:
beginalign
fracddxa_nx_n &=a_nnx^n-1\
fracddxa_nnx^n-1 & = a_nn(n-1)x^n-2\
fracddxa_nn(n-1)x^n-2& =a_nn(n-1)(n-2)x^n-3
\ &vdots
\ fracddxa_nn(n-1)(n-2)cdots(3)x^2& =a_nn(n-1)(n-2)cdots(3)(2)x
\ fracddxa_nn(n-1)(n-2)cdots(3)(2)x&=a_nn(n-1)(n-2)cdots(3)(2)(1)
\ & = a_ncdot n!
endalign
answered Jul 30 at 1:10
高田航
1,116318
answered Jul 30 at 1:10
高田航
1,116318
answered Jul 30 at 1:10
高田航
1,116318
answered Jul 30 at 1:10
高田航
1,116318
1,116318
add a comment |Â
add a comment |Â
up vote
0
down vote
First, you should convince yourself that all the terms on the right after the first do not contribute because you eventually take the derivative of a constant, which is zero.
Then do it by hand for $n=3$. We want the third derivative of $a_3x^3$. The first derivative is $3a_3x^2$, the second is $3cdot 2 a_3x$, the third is $3cdot 2 cdot 1a_3=3!a_3$
Now convince yourself that this is general and $f^n(x)=a_nn!$
Your proposed answer has two problems: the exponent decreases by $1$ each time you take a derivative so the second one should be $n-1$ and there are $n$ derivatives, not $2$. The numbers you bring down are multiplied, not added.
answered Jul 30 at 1:00
Ross Millikan
275k21185351
add a comment |Â
up vote
0
down vote
First, you should convince yourself that all the terms on the right after the first do not contribute because you eventually take the derivative of a constant, which is zero.
Then do it by hand for $n=3$. We want the third derivative of $a_3x^3$. The first derivative is $3a_3x^2$, the second is $3cdot 2 a_3x$, the third is $3cdot 2 cdot 1a_3=3!a_3$
Now convince yourself that this is general and $f^n(x)=a_nn!$
Your proposed answer has two problems: the exponent decreases by $1$ each time you take a derivative so the second one should be $n-1$ and there are $n$ derivatives, not $2$. The numbers you bring down are multiplied, not added.
answered Jul 30 at 1:00
Ross Millikan
275k21185351
add a comment |Â
up vote
0
down vote
up vote
0
down vote
First, you should convince yourself that all the terms on the right after the first do not contribute because you eventually take the derivative of a constant, which is zero.
Then do it by hand for $n=3$. We want the third derivative of $a_3x^3$. The first derivative is $3a_3x^2$, the second is $3cdot 2 a_3x$, the third is $3cdot 2 cdot 1a_3=3!a_3$
Now convince yourself that this is general and $f^n(x)=a_nn!$
Your proposed answer has two problems: the exponent decreases by $1$ each time you take a derivative so the second one should be $n-1$ and there are $n$ derivatives, not $2$. The numbers you bring down are multiplied, not added.
answered Jul 30 at 1:00
Ross Millikan
275k21185351
First, you should convince yourself that all the terms on the right after the first do not contribute because you eventually take the derivative of a constant, which is zero.
Then do it by hand for $n=3$. We want the third derivative of $a_3x^3$. The first derivative is $3a_3x^2$, the second is $3cdot 2 a_3x$, the third is $3cdot 2 cdot 1a_3=3!a_3$
Now convince yourself that this is general and $f^n(x)=a_nn!$
Your proposed answer has two problems: the exponent decreases by $1$ each time you take a derivative so the second one should be $n-1$ and there are $n$ derivatives, not $2$. The numbers you bring down are multiplied, not added.
answered Jul 30 at 1:00
Ross Millikan
275k21185351
answered Jul 30 at 1:00
Ross Millikan
275k21185351
answered Jul 30 at 1:00
Ross Millikan
275k21185351
answered Jul 30 at 1:00
Ross Millikan
275k21185351
275k21185351
add a comment |Â
add a comment |Â
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1
Start with $,left(x^nright)' = n x^n-1,$, $,left(x^nright)'' = n(n-1) x^n-2,$, $,left(x^nright)''' = n(n-1)(n-2) x^n-3 ,ldots,$
– dxiv
Jul 30 at 0:34
Hint: $(x^n)' = n x^n-1$, but $(nx^n-1)' = n(n-1) x^n-2$, and so on. Repeated application of the power rule give a falling factorial coefficient.
– Xander Henderson
Jul 30 at 0:35
The exercise book should say the answer is $a_nn(n-1)...1$.
– é«˜ç”°èˆª
Jul 30 at 0:39
You are right, that is the answer in the book. I badly transcribed it, I modified my question. Thank you
– Cedric Martens
Jul 30 at 0:41