Derivative of constants? (Applications of Integrals)
$begingroup$
I'm working on a list of problems from a chapter titled "Applications of Integrals" (area, volume, work etc.) and I have this problem:
$f(x)=(2+frac{9}{2})^3$ find $f'(x)$
As far as I know, this notation refers to the derivative of $f(x)$, yet this just be zero since it is only constants. I'm guessing there is more to this problem than that considering that it's surrounded by integrals on my worksheet, so am I missing something? Unless $f'(x)$ does not refer to a derivative? Thanks very much in advance.
calculus
asked Dec 7 '18 at 22:50
JSmithJSmith
224
$endgroup$
add a comment |
$begingroup$
I'm working on a list of problems from a chapter titled "Applications of Integrals" (area, volume, work etc.) and I have this problem:
$f(x)=(2+frac{9}{2})^3$ find $f'(x)$
As far as I know, this notation refers to the derivative of $f(x)$, yet this just be zero since it is only constants. I'm guessing there is more to this problem than that considering that it's surrounded by integrals on my worksheet, so am I missing something? Unless $f'(x)$ does not refer to a derivative? Thanks very much in advance.
calculus
asked Dec 7 '18 at 22:50
JSmithJSmith
224
$endgroup$
$begingroup$
" I'm guessing there is more to this problem than that " Nope, there isn't.
$endgroup$
– fleablood
Dec 7 '18 at 22:53
1
$begingroup$
Maybe it was a typo?
$endgroup$
– TrivialCase
Dec 7 '18 at 22:55
add a comment |
$begingroup$
I'm working on a list of problems from a chapter titled "Applications of Integrals" (area, volume, work etc.) and I have this problem:
$f(x)=(2+frac{9}{2})^3$ find $f'(x)$
As far as I know, this notation refers to the derivative of $f(x)$, yet this just be zero since it is only constants. I'm guessing there is more to this problem than that considering that it's surrounded by integrals on my worksheet, so am I missing something? Unless $f'(x)$ does not refer to a derivative? Thanks very much in advance.
calculus
asked Dec 7 '18 at 22:50
JSmithJSmith
224
$endgroup$
I'm working on a list of problems from a chapter titled "Applications of Integrals" (area, volume, work etc.) and I have this problem:
$f(x)=(2+frac{9}{2})^3$ find $f'(x)$
As far as I know, this notation refers to the derivative of $f(x)$, yet this just be zero since it is only constants. I'm guessing there is more to this problem than that considering that it's surrounded by integrals on my worksheet, so am I missing something? Unless $f'(x)$ does not refer to a derivative? Thanks very much in advance.
calculus
calculus
asked Dec 7 '18 at 22:50
JSmithJSmith
224
asked Dec 7 '18 at 22:50
JSmithJSmith
224
asked Dec 7 '18 at 22:50
JSmithJSmith
224
asked Dec 7 '18 at 22:50
JSmithJSmith
224
asked Dec 7 '18 at 22:50
JSmithJSmith
224
224
$begingroup$
" I'm guessing there is more to this problem than that " Nope, there isn't.
$endgroup$
– fleablood
Dec 7 '18 at 22:53
1
$begingroup$
Maybe it was a typo?
$endgroup$
– TrivialCase
Dec 7 '18 at 22:55
add a comment |
$begingroup$
" I'm guessing there is more to this problem than that " Nope, there isn't.
$endgroup$
– fleablood
Dec 7 '18 at 22:53
1
$begingroup$
Maybe it was a typo?
$endgroup$
– TrivialCase
Dec 7 '18 at 22:55
$begingroup$
" I'm guessing there is more to this problem than that " Nope, there isn't.
$endgroup$
– fleablood
Dec 7 '18 at 22:53
$begingroup$
" I'm guessing there is more to this problem than that " Nope, there isn't.
$endgroup$
– fleablood
Dec 7 '18 at 22:53
1
1
$begingroup$
Maybe it was a typo?
$endgroup$
– TrivialCase
Dec 7 '18 at 22:55
$begingroup$
Maybe it was a typo?
$endgroup$
– TrivialCase
Dec 7 '18 at 22:55
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I suspect the book made a typo and there is supposed to be an $x$ in there somewhere.
Or maybe the book wants you to prove that all the rules of derivatives still hold.
By chain rule if $j(x) = 2; g(x) = x + frac 92; h(x) = x^3$ so $f(x) = h(g(j(x)))$ then
$j'(x) = 0; g'(x) = 1; h'(x) = 3x^2$ so by chain rule:
$f'(x)= [h(g(j(x)))]' = h'(g(j(x)))g'(j(x))h'(x) = 3(2+frac 92)^2cdot 1cdot 0 = 0$
But that was silly.
I bet it was a typo.
====
But obviously you are right $f(x) = (2 +frac 92)^3 = (frac {13}2)^3 = 2460.375$ is a constant so $f'(x) = 0$.
answered Dec 7 '18 at 22:57
fleabloodfleablood
70.3k22685
$endgroup$
$begingroup$
Hmm, I was thinking something similar about the chain rule, but yeah it still results in zero. All right, I'll be sure to ask around on Monday, I appreciate the explanation, though.
$endgroup$
– JSmith
Dec 7 '18 at 23:03
$begingroup$
It's not a bad exercise to prove that different rules will still apply when attack by different methods. It's a bit excessive to apply the chain rule to everything but you can.
$endgroup$
– fleablood
Dec 7 '18 at 23:05
$begingroup$
Yet another option: trick question designed to make you get distracted by the expression and do lots of unnecessary work! "But it's just a constant" might be the intended correct answer.
$endgroup$
– timtfj
Dec 8 '18 at 0:24
$begingroup$
Yes, that is actually more likely. And it is .... just a constant.
$endgroup$
– fleablood
Dec 8 '18 at 1:21
add a comment |
$begingroup$
The only two possibilities I can think of is that there is a variable you have misread as a number, or $f$ is elsewhere defined in terms of some expression, and you're being asked to find what its derivative is when that expression is equal to $(2+frac92)^2$. Normally, images are discouraged, but one might help in this case.
answered Dec 7 '18 at 22:56
AcccumulationAcccumulation
6,9042618
$endgroup$
$begingroup$
THird option. The book made a typo.
$endgroup$
– fleablood
Dec 7 '18 at 23:00
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I suspect the book made a typo and there is supposed to be an $x$ in there somewhere.
Or maybe the book wants you to prove that all the rules of derivatives still hold.
By chain rule if $j(x) = 2; g(x) = x + frac 92; h(x) = x^3$ so $f(x) = h(g(j(x)))$ then
$j'(x) = 0; g'(x) = 1; h'(x) = 3x^2$ so by chain rule:
$f'(x)= [h(g(j(x)))]' = h'(g(j(x)))g'(j(x))h'(x) = 3(2+frac 92)^2cdot 1cdot 0 = 0$
But that was silly.
I bet it was a typo.
====
But obviously you are right $f(x) = (2 +frac 92)^3 = (frac {13}2)^3 = 2460.375$ is a constant so $f'(x) = 0$.
answered Dec 7 '18 at 22:57
fleabloodfleablood
70.3k22685
$endgroup$
$begingroup$
Hmm, I was thinking something similar about the chain rule, but yeah it still results in zero. All right, I'll be sure to ask around on Monday, I appreciate the explanation, though.
$endgroup$
– JSmith
Dec 7 '18 at 23:03
$begingroup$
It's not a bad exercise to prove that different rules will still apply when attack by different methods. It's a bit excessive to apply the chain rule to everything but you can.
$endgroup$
– fleablood
Dec 7 '18 at 23:05
$begingroup$
Yet another option: trick question designed to make you get distracted by the expression and do lots of unnecessary work! "But it's just a constant" might be the intended correct answer.
$endgroup$
– timtfj
Dec 8 '18 at 0:24
$begingroup$
Yes, that is actually more likely. And it is .... just a constant.
$endgroup$
– fleablood
Dec 8 '18 at 1:21
add a comment |
$begingroup$
I suspect the book made a typo and there is supposed to be an $x$ in there somewhere.
Or maybe the book wants you to prove that all the rules of derivatives still hold.
By chain rule if $j(x) = 2; g(x) = x + frac 92; h(x) = x^3$ so $f(x) = h(g(j(x)))$ then
$j'(x) = 0; g'(x) = 1; h'(x) = 3x^2$ so by chain rule:
$f'(x)= [h(g(j(x)))]' = h'(g(j(x)))g'(j(x))h'(x) = 3(2+frac 92)^2cdot 1cdot 0 = 0$
But that was silly.
I bet it was a typo.
====
But obviously you are right $f(x) = (2 +frac 92)^3 = (frac {13}2)^3 = 2460.375$ is a constant so $f'(x) = 0$.
answered Dec 7 '18 at 22:57
fleabloodfleablood
70.3k22685
$endgroup$
$begingroup$
Hmm, I was thinking something similar about the chain rule, but yeah it still results in zero. All right, I'll be sure to ask around on Monday, I appreciate the explanation, though.
$endgroup$
– JSmith
Dec 7 '18 at 23:03
$begingroup$
It's not a bad exercise to prove that different rules will still apply when attack by different methods. It's a bit excessive to apply the chain rule to everything but you can.
$endgroup$
– fleablood
Dec 7 '18 at 23:05
$begingroup$
Yet another option: trick question designed to make you get distracted by the expression and do lots of unnecessary work! "But it's just a constant" might be the intended correct answer.
$endgroup$
– timtfj
Dec 8 '18 at 0:24
$begingroup$
Yes, that is actually more likely. And it is .... just a constant.
$endgroup$
– fleablood
Dec 8 '18 at 1:21
add a comment |
$begingroup$
I suspect the book made a typo and there is supposed to be an $x$ in there somewhere.
Or maybe the book wants you to prove that all the rules of derivatives still hold.
By chain rule if $j(x) = 2; g(x) = x + frac 92; h(x) = x^3$ so $f(x) = h(g(j(x)))$ then
$j'(x) = 0; g'(x) = 1; h'(x) = 3x^2$ so by chain rule:
$f'(x)= [h(g(j(x)))]' = h'(g(j(x)))g'(j(x))h'(x) = 3(2+frac 92)^2cdot 1cdot 0 = 0$
But that was silly.
I bet it was a typo.
====
But obviously you are right $f(x) = (2 +frac 92)^3 = (frac {13}2)^3 = 2460.375$ is a constant so $f'(x) = 0$.
answered Dec 7 '18 at 22:57
fleabloodfleablood
70.3k22685
$endgroup$
I suspect the book made a typo and there is supposed to be an $x$ in there somewhere.
Or maybe the book wants you to prove that all the rules of derivatives still hold.
By chain rule if $j(x) = 2; g(x) = x + frac 92; h(x) = x^3$ so $f(x) = h(g(j(x)))$ then
$j'(x) = 0; g'(x) = 1; h'(x) = 3x^2$ so by chain rule:
$f'(x)= [h(g(j(x)))]' = h'(g(j(x)))g'(j(x))h'(x) = 3(2+frac 92)^2cdot 1cdot 0 = 0$
But that was silly.
I bet it was a typo.
====
But obviously you are right $f(x) = (2 +frac 92)^3 = (frac {13}2)^3 = 2460.375$ is a constant so $f'(x) = 0$.
answered Dec 7 '18 at 22:57
fleabloodfleablood
70.3k22685
edited Dec 7 '18 at 23:04
answered Dec 7 '18 at 22:57
fleabloodfleablood
70.3k22685
answered Dec 7 '18 at 22:57
fleabloodfleablood
70.3k22685
answered Dec 7 '18 at 22:57
fleabloodfleablood
70.3k22685
70.3k22685
$begingroup$
Hmm, I was thinking something similar about the chain rule, but yeah it still results in zero. All right, I'll be sure to ask around on Monday, I appreciate the explanation, though.
$endgroup$
– JSmith
Dec 7 '18 at 23:03
$begingroup$
It's not a bad exercise to prove that different rules will still apply when attack by different methods. It's a bit excessive to apply the chain rule to everything but you can.
$endgroup$
– fleablood
Dec 7 '18 at 23:05
$begingroup$
Yet another option: trick question designed to make you get distracted by the expression and do lots of unnecessary work! "But it's just a constant" might be the intended correct answer.
$endgroup$
– timtfj
Dec 8 '18 at 0:24
$begingroup$
Yes, that is actually more likely. And it is .... just a constant.
$endgroup$
– fleablood
Dec 8 '18 at 1:21
add a comment |
$begingroup$
Hmm, I was thinking something similar about the chain rule, but yeah it still results in zero. All right, I'll be sure to ask around on Monday, I appreciate the explanation, though.
$endgroup$
– JSmith
Dec 7 '18 at 23:03
$begingroup$
It's not a bad exercise to prove that different rules will still apply when attack by different methods. It's a bit excessive to apply the chain rule to everything but you can.
$endgroup$
– fleablood
Dec 7 '18 at 23:05
$begingroup$
Yet another option: trick question designed to make you get distracted by the expression and do lots of unnecessary work! "But it's just a constant" might be the intended correct answer.
$endgroup$
– timtfj
Dec 8 '18 at 0:24
$begingroup$
Yes, that is actually more likely. And it is .... just a constant.
$endgroup$
– fleablood
Dec 8 '18 at 1:21
$begingroup$
Hmm, I was thinking something similar about the chain rule, but yeah it still results in zero. All right, I'll be sure to ask around on Monday, I appreciate the explanation, though.
$endgroup$
– JSmith
Dec 7 '18 at 23:03
$begingroup$
Hmm, I was thinking something similar about the chain rule, but yeah it still results in zero. All right, I'll be sure to ask around on Monday, I appreciate the explanation, though.
$endgroup$
– JSmith
Dec 7 '18 at 23:03
$begingroup$
It's not a bad exercise to prove that different rules will still apply when attack by different methods. It's a bit excessive to apply the chain rule to everything but you can.
$endgroup$
– fleablood
Dec 7 '18 at 23:05
$begingroup$
It's not a bad exercise to prove that different rules will still apply when attack by different methods. It's a bit excessive to apply the chain rule to everything but you can.
$endgroup$
– fleablood
Dec 7 '18 at 23:05
$begingroup$
Yet another option: trick question designed to make you get distracted by the expression and do lots of unnecessary work! "But it's just a constant" might be the intended correct answer.
$endgroup$
– timtfj
Dec 8 '18 at 0:24
$begingroup$
Yet another option: trick question designed to make you get distracted by the expression and do lots of unnecessary work! "But it's just a constant" might be the intended correct answer.
$endgroup$
– timtfj
Dec 8 '18 at 0:24
$begingroup$
Yes, that is actually more likely. And it is .... just a constant.
$endgroup$
– fleablood
Dec 8 '18 at 1:21
$begingroup$
Yes, that is actually more likely. And it is .... just a constant.
$endgroup$
– fleablood
Dec 8 '18 at 1:21
add a comment |
$begingroup$
The only two possibilities I can think of is that there is a variable you have misread as a number, or $f$ is elsewhere defined in terms of some expression, and you're being asked to find what its derivative is when that expression is equal to $(2+frac92)^2$. Normally, images are discouraged, but one might help in this case.
answered Dec 7 '18 at 22:56
AcccumulationAcccumulation
6,9042618
$endgroup$
$begingroup$
THird option. The book made a typo.
$endgroup$
– fleablood
Dec 7 '18 at 23:00
add a comment |
$begingroup$
The only two possibilities I can think of is that there is a variable you have misread as a number, or $f$ is elsewhere defined in terms of some expression, and you're being asked to find what its derivative is when that expression is equal to $(2+frac92)^2$. Normally, images are discouraged, but one might help in this case.
answered Dec 7 '18 at 22:56
AcccumulationAcccumulation
6,9042618
$endgroup$
$begingroup$
THird option. The book made a typo.
$endgroup$
– fleablood
Dec 7 '18 at 23:00
add a comment |
$begingroup$
The only two possibilities I can think of is that there is a variable you have misread as a number, or $f$ is elsewhere defined in terms of some expression, and you're being asked to find what its derivative is when that expression is equal to $(2+frac92)^2$. Normally, images are discouraged, but one might help in this case.
answered Dec 7 '18 at 22:56
AcccumulationAcccumulation
6,9042618
$endgroup$
The only two possibilities I can think of is that there is a variable you have misread as a number, or $f$ is elsewhere defined in terms of some expression, and you're being asked to find what its derivative is when that expression is equal to $(2+frac92)^2$. Normally, images are discouraged, but one might help in this case.
answered Dec 7 '18 at 22:56
AcccumulationAcccumulation
6,9042618
answered Dec 7 '18 at 22:56
AcccumulationAcccumulation
6,9042618
answered Dec 7 '18 at 22:56
AcccumulationAcccumulation
6,9042618
answered Dec 7 '18 at 22:56
AcccumulationAcccumulation
6,9042618
6,9042618
$begingroup$
THird option. The book made a typo.
$endgroup$
– fleablood
Dec 7 '18 at 23:00
add a comment |
$begingroup$
THird option. The book made a typo.
$endgroup$
– fleablood
Dec 7 '18 at 23:00
$begingroup$
THird option. The book made a typo.
$endgroup$
– fleablood
Dec 7 '18 at 23:00
$begingroup$
THird option. The book made a typo.
$endgroup$
– fleablood
Dec 7 '18 at 23:00
add a comment |
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$begingroup$
" I'm guessing there is more to this problem than that " Nope, there isn't.
$endgroup$
– fleablood
Dec 7 '18 at 22:53
1
$begingroup$
Maybe it was a typo?
$endgroup$
– TrivialCase
Dec 7 '18 at 22:55