How is the notation $frac{d}{dx} (f^3)(1)$ interpreted?
up vote
4
down vote
favorite
How is the following notation interpreted?
$$frac{d}{dx} (f^3)(1)$$
Does this evaluate to $3cdot f(1)^2cdot f'(1) $, or is it simply the derivative of a constant and equal to 0?
calculus derivatives
edited Nov 12 at 16:12
user587192
1,410112
asked Nov 12 at 1:21
Is12Prime
12119
|
show 2 more comments
up vote
4
down vote
favorite
How is the following notation interpreted?
$$frac{d}{dx} (f^3)(1)$$
Does this evaluate to $3cdot f(1)^2cdot f'(1) $, or is it simply the derivative of a constant and equal to 0?
calculus derivatives
edited Nov 12 at 16:12
user587192
1,410112
asked Nov 12 at 1:21
Is12Prime
12119
1
Where did you get it from? That notation looks very strange.
– Mike R.
Nov 12 at 1:27
It was in my textbook
– Is12Prime
Nov 12 at 1:27
Honestly, it depends on the context...the $3$ could mean the third derivative, or the function cubed.
– Rushabh Mehta
Nov 12 at 1:30
2
In that case, I'd cautiously vouch for your evaluation.
– Rushabh Mehta
Nov 12 at 1:31
1
I consider this notation as really clean and unambiguous! It is as you’ve written in the first place, the derivative of $f^3$ evaluated in $1$.
– Sebastian Bechtel
Nov 12 at 8:19
|
show 2 more comments
up vote
4
down vote
favorite
up vote
4
down vote
favorite
How is the following notation interpreted?
$$frac{d}{dx} (f^3)(1)$$
Does this evaluate to $3cdot f(1)^2cdot f'(1) $, or is it simply the derivative of a constant and equal to 0?
calculus derivatives
edited Nov 12 at 16:12
user587192
1,410112
asked Nov 12 at 1:21
Is12Prime
12119
How is the following notation interpreted?
$$frac{d}{dx} (f^3)(1)$$
Does this evaluate to $3cdot f(1)^2cdot f'(1) $, or is it simply the derivative of a constant and equal to 0?
calculus derivatives
calculus derivatives
edited Nov 12 at 16:12
user587192
1,410112
asked Nov 12 at 1:21
Is12Prime
12119
edited Nov 12 at 16:12
user587192
1,410112
asked Nov 12 at 1:21
Is12Prime
12119
edited Nov 12 at 16:12
user587192
1,410112
edited Nov 12 at 16:12
user587192
1,410112
edited Nov 12 at 16:12
user587192
1,410112
1,410112
asked Nov 12 at 1:21
Is12Prime
12119
asked Nov 12 at 1:21
Is12Prime
12119
asked Nov 12 at 1:21
Is12Prime
12119
12119
1
Where did you get it from? That notation looks very strange.
– Mike R.
Nov 12 at 1:27
It was in my textbook
– Is12Prime
Nov 12 at 1:27
Honestly, it depends on the context...the $3$ could mean the third derivative, or the function cubed.
– Rushabh Mehta
Nov 12 at 1:30
2
In that case, I'd cautiously vouch for your evaluation.
– Rushabh Mehta
Nov 12 at 1:31
1
I consider this notation as really clean and unambiguous! It is as you’ve written in the first place, the derivative of $f^3$ evaluated in $1$.
– Sebastian Bechtel
Nov 12 at 8:19
|
show 2 more comments
1
Where did you get it from? That notation looks very strange.
– Mike R.
Nov 12 at 1:27
It was in my textbook
– Is12Prime
Nov 12 at 1:27
Honestly, it depends on the context...the $3$ could mean the third derivative, or the function cubed.
– Rushabh Mehta
Nov 12 at 1:30
2
In that case, I'd cautiously vouch for your evaluation.
– Rushabh Mehta
Nov 12 at 1:31
1
I consider this notation as really clean and unambiguous! It is as you’ve written in the first place, the derivative of $f^3$ evaluated in $1$.
– Sebastian Bechtel
Nov 12 at 8:19
1
1
Where did you get it from? That notation looks very strange.
– Mike R.
Nov 12 at 1:27
Where did you get it from? That notation looks very strange.
– Mike R.
Nov 12 at 1:27
It was in my textbook
– Is12Prime
Nov 12 at 1:27
It was in my textbook
– Is12Prime
Nov 12 at 1:27
Honestly, it depends on the context...the $3$ could mean the third derivative, or the function cubed.
– Rushabh Mehta
Nov 12 at 1:30
Honestly, it depends on the context...the $3$ could mean the third derivative, or the function cubed.
– Rushabh Mehta
Nov 12 at 1:30
2
2
In that case, I'd cautiously vouch for your evaluation.
– Rushabh Mehta
Nov 12 at 1:31
In that case, I'd cautiously vouch for your evaluation.
– Rushabh Mehta
Nov 12 at 1:31
1
1
I consider this notation as really clean and unambiguous! It is as you’ve written in the first place, the derivative of $f^3$ evaluated in $1$.
– Sebastian Bechtel
Nov 12 at 8:19
I consider this notation as really clean and unambiguous! It is as you’ve written in the first place, the derivative of $f^3$ evaluated in $1$.
– Sebastian Bechtel
Nov 12 at 8:19
|
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
8
down vote
accepted
I'd say it probably means the derivative of the function $f$ (whatever that function happens to be) cubed evaluated at $1$. And I'd also suggest that it would probably be better to denote $f$ as $f(x)$:
$$
frac{d}{dx}(f^3)(1) = 3cdot [f(x)]^2cdot f'(x)vert_{x=1}=3cdot [f(1)]^2cdot f'(1)
$$
answered Nov 12 at 1:32
Mike R.
1,339212
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
I'd say it probably means the derivative of the function $f$ (whatever that function happens to be) cubed evaluated at $1$. And I'd also suggest that it would probably be better to denote $f$ as $f(x)$:
$$
frac{d}{dx}(f^3)(1) = 3cdot [f(x)]^2cdot f'(x)vert_{x=1}=3cdot [f(1)]^2cdot f'(1)
$$
answered Nov 12 at 1:32
Mike R.
1,339212
add a comment |
up vote
8
down vote
accepted
I'd say it probably means the derivative of the function $f$ (whatever that function happens to be) cubed evaluated at $1$. And I'd also suggest that it would probably be better to denote $f$ as $f(x)$:
$$
frac{d}{dx}(f^3)(1) = 3cdot [f(x)]^2cdot f'(x)vert_{x=1}=3cdot [f(1)]^2cdot f'(1)
$$
answered Nov 12 at 1:32
Mike R.
1,339212
add a comment |
up vote
8
down vote
accepted
up vote
8
down vote
accepted
I'd say it probably means the derivative of the function $f$ (whatever that function happens to be) cubed evaluated at $1$. And I'd also suggest that it would probably be better to denote $f$ as $f(x)$:
$$
frac{d}{dx}(f^3)(1) = 3cdot [f(x)]^2cdot f'(x)vert_{x=1}=3cdot [f(1)]^2cdot f'(1)
$$
answered Nov 12 at 1:32
Mike R.
1,339212
I'd say it probably means the derivative of the function $f$ (whatever that function happens to be) cubed evaluated at $1$. And I'd also suggest that it would probably be better to denote $f$ as $f(x)$:
$$
frac{d}{dx}(f^3)(1) = 3cdot [f(x)]^2cdot f'(x)vert_{x=1}=3cdot [f(1)]^2cdot f'(1)
$$
answered Nov 12 at 1:32
Mike R.
1,339212
edited Nov 12 at 1:39
answered Nov 12 at 1:32
Mike R.
1,339212
answered Nov 12 at 1:32
Mike R.
1,339212
answered Nov 12 at 1:32
Mike R.
1,339212
1,339212
add a comment |
add a comment |
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1
Where did you get it from? That notation looks very strange.
– Mike R.
Nov 12 at 1:27
It was in my textbook
– Is12Prime
Nov 12 at 1:27
Honestly, it depends on the context...the $3$ could mean the third derivative, or the function cubed.
– Rushabh Mehta
Nov 12 at 1:30
2
In that case, I'd cautiously vouch for your evaluation.
– Rushabh Mehta
Nov 12 at 1:31
1
I consider this notation as really clean and unambiguous! It is as you’ve written in the first place, the derivative of $f^3$ evaluated in $1$.
– Sebastian Bechtel
Nov 12 at 8:19