How to get this formula: $(f^2(x))'=2f(x)cdot f'(x)$? [closed]
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$$(f^2(x))'=2f(x)cdot f'(x)$$
What is it ?
How can I get this formula.
Thank you for help. And what is the name of this formula ?
derivatives
edited Nov 26 at 16:59
Martin Sleziak
44.5k7115269
asked Nov 26 at 16:23
Alexey
62
closed as off-topic by John B, Brahadeesh, Cyclohexanol., user302797, A. Pongrácz Nov 27 at 10:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John B, Brahadeesh, Cyclohexanol., user302797, A. Pongrácz
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
-5
down vote
favorite
$$(f^2(x))'=2f(x)cdot f'(x)$$
What is it ?
How can I get this formula.
Thank you for help. And what is the name of this formula ?
derivatives
edited Nov 26 at 16:59
Martin Sleziak
44.5k7115269
asked Nov 26 at 16:23
Alexey
62
closed as off-topic by John B, Brahadeesh, Cyclohexanol., user302797, A. Pongrácz Nov 27 at 10:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John B, Brahadeesh, Cyclohexanol., user302797, A. Pongrácz
If this question can be reworded to fit the rules in the help center, please edit the question.
1
Chain rule?${}$
– MisterRiemann
Nov 26 at 16:24
2
Or just the product rule applies to $f(x)times f(x)$.
– lulu
Nov 26 at 16:25
With convention $f^2(x)=(f(x))^2$, we apply chain rule.
– Yadati Kiran
Nov 26 at 16:40
add a comment |
up vote
-5
down vote
favorite
up vote
-5
down vote
favorite
$$(f^2(x))'=2f(x)cdot f'(x)$$
What is it ?
How can I get this formula.
Thank you for help. And what is the name of this formula ?
derivatives
edited Nov 26 at 16:59
Martin Sleziak
44.5k7115269
asked Nov 26 at 16:23
Alexey
62
$$(f^2(x))'=2f(x)cdot f'(x)$$
What is it ?
How can I get this formula.
Thank you for help. And what is the name of this formula ?
derivatives
derivatives
edited Nov 26 at 16:59
Martin Sleziak
44.5k7115269
asked Nov 26 at 16:23
Alexey
62
edited Nov 26 at 16:59
Martin Sleziak
44.5k7115269
asked Nov 26 at 16:23
Alexey
62
edited Nov 26 at 16:59
Martin Sleziak
44.5k7115269
edited Nov 26 at 16:59
Martin Sleziak
44.5k7115269
edited Nov 26 at 16:59
Martin Sleziak
44.5k7115269
44.5k7115269
asked Nov 26 at 16:23
Alexey
62
asked Nov 26 at 16:23
Alexey
62
asked Nov 26 at 16:23
Alexey
62
62
closed as off-topic by John B, Brahadeesh, Cyclohexanol., user302797, A. Pongrácz Nov 27 at 10:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John B, Brahadeesh, Cyclohexanol., user302797, A. Pongrácz
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by John B, Brahadeesh, Cyclohexanol., user302797, A. Pongrácz Nov 27 at 10:13
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – John B, Brahadeesh, Cyclohexanol., user302797, A. Pongrácz
If this question can be reworded to fit the rules in the help center, please edit the question.
1
Chain rule?${}$
– MisterRiemann
Nov 26 at 16:24
2
Or just the product rule applies to $f(x)times f(x)$.
– lulu
Nov 26 at 16:25
With convention $f^2(x)=(f(x))^2$, we apply chain rule.
– Yadati Kiran
Nov 26 at 16:40
add a comment |
1
Chain rule?${}$
– MisterRiemann
Nov 26 at 16:24
2
Or just the product rule applies to $f(x)times f(x)$.
– lulu
Nov 26 at 16:25
With convention $f^2(x)=(f(x))^2$, we apply chain rule.
– Yadati Kiran
Nov 26 at 16:40
1
1
Chain rule?${}$
– MisterRiemann
Nov 26 at 16:24
Chain rule?${}$
– MisterRiemann
Nov 26 at 16:24
2
2
Or just the product rule applies to $f(x)times f(x)$.
– lulu
Nov 26 at 16:25
Or just the product rule applies to $f(x)times f(x)$.
– lulu
Nov 26 at 16:25
With convention $f^2(x)=(f(x))^2$, we apply chain rule.
– Yadati Kiran
Nov 26 at 16:40
With convention $f^2(x)=(f(x))^2$, we apply chain rule.
– Yadati Kiran
Nov 26 at 16:40
add a comment |
1 Answer
1
active
oldest
votes
up vote
6
down vote
accepted
It's not a named rule in its own right. It's a special case of either the product rule, or the chain rule, depending on your personal taste.
Product rule:
$$
[f(x)cdot f(x)]' = f(x)cdot f'(x) + f'(x)cdot f(x)
$$
Chain rule: Take the function $g(x) = x^2$. Then $(f(x))^2 = g(f(x))$ and $g'(x) = 2x$. This gives
$$
[g(f(x))]' = g'(f(x))cdot f'(x) = 2f(x)cdot f'(x)
$$
answered Nov 26 at 16:25
Arthur
109k7103186
Thank you! You are my hero )
– Alexey
Nov 26 at 16:40
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
It's not a named rule in its own right. It's a special case of either the product rule, or the chain rule, depending on your personal taste.
Product rule:
$$
[f(x)cdot f(x)]' = f(x)cdot f'(x) + f'(x)cdot f(x)
$$
Chain rule: Take the function $g(x) = x^2$. Then $(f(x))^2 = g(f(x))$ and $g'(x) = 2x$. This gives
$$
[g(f(x))]' = g'(f(x))cdot f'(x) = 2f(x)cdot f'(x)
$$
answered Nov 26 at 16:25
Arthur
109k7103186
Thank you! You are my hero )
– Alexey
Nov 26 at 16:40
add a comment |
up vote
6
down vote
accepted
It's not a named rule in its own right. It's a special case of either the product rule, or the chain rule, depending on your personal taste.
Product rule:
$$
[f(x)cdot f(x)]' = f(x)cdot f'(x) + f'(x)cdot f(x)
$$
Chain rule: Take the function $g(x) = x^2$. Then $(f(x))^2 = g(f(x))$ and $g'(x) = 2x$. This gives
$$
[g(f(x))]' = g'(f(x))cdot f'(x) = 2f(x)cdot f'(x)
$$
answered Nov 26 at 16:25
Arthur
109k7103186
Thank you! You are my hero )
– Alexey
Nov 26 at 16:40
add a comment |
up vote
6
down vote
accepted
up vote
6
down vote
accepted
It's not a named rule in its own right. It's a special case of either the product rule, or the chain rule, depending on your personal taste.
Product rule:
$$
[f(x)cdot f(x)]' = f(x)cdot f'(x) + f'(x)cdot f(x)
$$
Chain rule: Take the function $g(x) = x^2$. Then $(f(x))^2 = g(f(x))$ and $g'(x) = 2x$. This gives
$$
[g(f(x))]' = g'(f(x))cdot f'(x) = 2f(x)cdot f'(x)
$$
answered Nov 26 at 16:25
Arthur
109k7103186
It's not a named rule in its own right. It's a special case of either the product rule, or the chain rule, depending on your personal taste.
Product rule:
$$
[f(x)cdot f(x)]' = f(x)cdot f'(x) + f'(x)cdot f(x)
$$
Chain rule: Take the function $g(x) = x^2$. Then $(f(x))^2 = g(f(x))$ and $g'(x) = 2x$. This gives
$$
[g(f(x))]' = g'(f(x))cdot f'(x) = 2f(x)cdot f'(x)
$$
answered Nov 26 at 16:25
Arthur
109k7103186
answered Nov 26 at 16:25
Arthur
109k7103186
answered Nov 26 at 16:25
Arthur
109k7103186
answered Nov 26 at 16:25
Arthur
109k7103186
109k7103186
Thank you! You are my hero )
– Alexey
Nov 26 at 16:40
add a comment |
Thank you! You are my hero )
– Alexey
Nov 26 at 16:40
Thank you! You are my hero )
– Alexey
Nov 26 at 16:40
Thank you! You are my hero )
– Alexey
Nov 26 at 16:40
add a comment |
1
Chain rule?${}$
– MisterRiemann
Nov 26 at 16:24
2
Or just the product rule applies to $f(x)times f(x)$.
– lulu
Nov 26 at 16:25
With convention $f^2(x)=(f(x))^2$, we apply chain rule.
– Yadati Kiran
Nov 26 at 16:40