What is derivative of $sin ax$ where $a$ is a constant?
up vote
1
down vote
favorite
What is the derivative of $sin a x$ where $a$ is a constant.
Actually, I'm studying Physics and not so well-versed with calculus. So, I have studied the basic rules of calculus but am stuck here.
I somewhat know about the product rule but don't get what to do if a constant is given in a trigonometric function, be it $sin ax $ or $cos ax$. Whatever..
Please help me get my concept clear.
Thank You!
calculus algebra-precalculus derivatives differential
edited Nov 29 at 19:19
Asaf Karagila♦
300k32421751
asked Nov 29 at 15:17
Chaku Daku
91
New contributor
Chaku Daku is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
1
down vote
favorite
What is the derivative of $sin a x$ where $a$ is a constant.
Actually, I'm studying Physics and not so well-versed with calculus. So, I have studied the basic rules of calculus but am stuck here.
I somewhat know about the product rule but don't get what to do if a constant is given in a trigonometric function, be it $sin ax $ or $cos ax$. Whatever..
Please help me get my concept clear.
Thank You!
calculus algebra-precalculus derivatives differential
edited Nov 29 at 19:19
Asaf Karagila♦
300k32421751
asked Nov 29 at 15:17
Chaku Daku
91
New contributor
Chaku Daku is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Apply the chain rule.
– user3482749
Nov 29 at 15:17
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
What is the derivative of $sin a x$ where $a$ is a constant.
Actually, I'm studying Physics and not so well-versed with calculus. So, I have studied the basic rules of calculus but am stuck here.
I somewhat know about the product rule but don't get what to do if a constant is given in a trigonometric function, be it $sin ax $ or $cos ax$. Whatever..
Please help me get my concept clear.
Thank You!
calculus algebra-precalculus derivatives differential
edited Nov 29 at 19:19
Asaf Karagila♦
300k32421751
asked Nov 29 at 15:17
Chaku Daku
91
New contributor
Chaku Daku is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What is the derivative of $sin a x$ where $a$ is a constant.
Actually, I'm studying Physics and not so well-versed with calculus. So, I have studied the basic rules of calculus but am stuck here.
I somewhat know about the product rule but don't get what to do if a constant is given in a trigonometric function, be it $sin ax $ or $cos ax$. Whatever..
Please help me get my concept clear.
Thank You!
calculus algebra-precalculus derivatives differential
calculus algebra-precalculus derivatives differential
edited Nov 29 at 19:19
Asaf Karagila♦
300k32421751
asked Nov 29 at 15:17
Chaku Daku
91
New contributor
Chaku Daku is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 29 at 19:19
Asaf Karagila♦
300k32421751
asked Nov 29 at 15:17
Chaku Daku
91
New contributor
Chaku Daku is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 29 at 19:19
Asaf Karagila♦
300k32421751
edited Nov 29 at 19:19
Asaf Karagila♦
300k32421751
edited Nov 29 at 19:19
Asaf Karagila♦
300k32421751
300k32421751
asked Nov 29 at 15:17
Chaku Daku
91
New contributor
Chaku Daku is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 29 at 15:17
Chaku Daku
91
asked Nov 29 at 15:17
Chaku Daku
91
91
New contributor
Chaku Daku is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Chaku Daku is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Chaku Daku is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Apply the chain rule.
– user3482749
Nov 29 at 15:17
add a comment |
1
Apply the chain rule.
– user3482749
Nov 29 at 15:17
1
1
Apply the chain rule.
– user3482749
Nov 29 at 15:17
Apply the chain rule.
– user3482749
Nov 29 at 15:17
add a comment |
3 Answers
3
active
oldest
votes
up vote
6
down vote
HINT
Recall that by chain rule
$$frac{d}{dx}[sin (f(x))]=cos (f(x))cdot f'(x)$$
answered Nov 29 at 15:20
gimusi
89.4k74495
add a comment |
up vote
3
down vote
begin{array}{c}
frac{d}{{dx}}left( {sin ax} right) = left( {cos ax} right)frac{d}{{dx}}left( {ax} right)\
= left( {cos ax} right) cdot a cdot frac{{dx}}{{dx}}\
= left( {cos ax} right) cdot a cdot 1\
= aleft( {cos ax} right)
end{array}
answered Nov 29 at 16:10
Krishna Srivastav
894
1
Out of interest why can you not just go straight from the first line to the last line without needing to have the $dx/dx$ part in there too?
– Chris
Nov 29 at 17:20
1
@Chris - because going straight from the first line to the last is simply skipping over steps that are required by a direct application of the theorems, but are so trivial people usually don't bother to write them down. Krishna Srivastav has simply chosen to show all the steps to make it clear how the full calculation proceeds. Occasionally a good teacher has to do this, as what makes those steps trivial is that everyone understands them. However, this is not always true of new students.
– Paul Sinclair
Nov 29 at 17:29
add a comment |
up vote
1
down vote
Derivative of $sin(ax) = a cos(ax)$ by Chain Rule.
edited Nov 29 at 15:42
Tianlalu
2,8811935
answered Nov 29 at 15:21
Rohit Bharadwaj
518
New contributor
Rohit Bharadwaj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
HINT
Recall that by chain rule
$$frac{d}{dx}[sin (f(x))]=cos (f(x))cdot f'(x)$$
answered Nov 29 at 15:20
gimusi
89.4k74495
add a comment |
up vote
6
down vote
HINT
Recall that by chain rule
$$frac{d}{dx}[sin (f(x))]=cos (f(x))cdot f'(x)$$
answered Nov 29 at 15:20
gimusi
89.4k74495
add a comment |
up vote
6
down vote
up vote
6
down vote
HINT
Recall that by chain rule
$$frac{d}{dx}[sin (f(x))]=cos (f(x))cdot f'(x)$$
answered Nov 29 at 15:20
gimusi
89.4k74495
HINT
Recall that by chain rule
$$frac{d}{dx}[sin (f(x))]=cos (f(x))cdot f'(x)$$
answered Nov 29 at 15:20
gimusi
89.4k74495
answered Nov 29 at 15:20
gimusi
89.4k74495
answered Nov 29 at 15:20
gimusi
89.4k74495
answered Nov 29 at 15:20
gimusi
89.4k74495
89.4k74495
add a comment |
add a comment |
up vote
3
down vote
begin{array}{c}
frac{d}{{dx}}left( {sin ax} right) = left( {cos ax} right)frac{d}{{dx}}left( {ax} right)\
= left( {cos ax} right) cdot a cdot frac{{dx}}{{dx}}\
= left( {cos ax} right) cdot a cdot 1\
= aleft( {cos ax} right)
end{array}
answered Nov 29 at 16:10
Krishna Srivastav
894
1
Out of interest why can you not just go straight from the first line to the last line without needing to have the $dx/dx$ part in there too?
– Chris
Nov 29 at 17:20
1
@Chris - because going straight from the first line to the last is simply skipping over steps that are required by a direct application of the theorems, but are so trivial people usually don't bother to write them down. Krishna Srivastav has simply chosen to show all the steps to make it clear how the full calculation proceeds. Occasionally a good teacher has to do this, as what makes those steps trivial is that everyone understands them. However, this is not always true of new students.
– Paul Sinclair
Nov 29 at 17:29
add a comment |
up vote
3
down vote
begin{array}{c}
frac{d}{{dx}}left( {sin ax} right) = left( {cos ax} right)frac{d}{{dx}}left( {ax} right)\
= left( {cos ax} right) cdot a cdot frac{{dx}}{{dx}}\
= left( {cos ax} right) cdot a cdot 1\
= aleft( {cos ax} right)
end{array}
answered Nov 29 at 16:10
Krishna Srivastav
894
1
Out of interest why can you not just go straight from the first line to the last line without needing to have the $dx/dx$ part in there too?
– Chris
Nov 29 at 17:20
1
@Chris - because going straight from the first line to the last is simply skipping over steps that are required by a direct application of the theorems, but are so trivial people usually don't bother to write them down. Krishna Srivastav has simply chosen to show all the steps to make it clear how the full calculation proceeds. Occasionally a good teacher has to do this, as what makes those steps trivial is that everyone understands them. However, this is not always true of new students.
– Paul Sinclair
Nov 29 at 17:29
add a comment |
up vote
3
down vote
up vote
3
down vote
begin{array}{c}
frac{d}{{dx}}left( {sin ax} right) = left( {cos ax} right)frac{d}{{dx}}left( {ax} right)\
= left( {cos ax} right) cdot a cdot frac{{dx}}{{dx}}\
= left( {cos ax} right) cdot a cdot 1\
= aleft( {cos ax} right)
end{array}
answered Nov 29 at 16:10
Krishna Srivastav
894
begin{array}{c}
frac{d}{{dx}}left( {sin ax} right) = left( {cos ax} right)frac{d}{{dx}}left( {ax} right)\
= left( {cos ax} right) cdot a cdot frac{{dx}}{{dx}}\
= left( {cos ax} right) cdot a cdot 1\
= aleft( {cos ax} right)
end{array}
answered Nov 29 at 16:10
Krishna Srivastav
894
answered Nov 29 at 16:10
Krishna Srivastav
894
answered Nov 29 at 16:10
Krishna Srivastav
894
answered Nov 29 at 16:10
Krishna Srivastav
894
894
1
Out of interest why can you not just go straight from the first line to the last line without needing to have the $dx/dx$ part in there too?
– Chris
Nov 29 at 17:20
1
@Chris - because going straight from the first line to the last is simply skipping over steps that are required by a direct application of the theorems, but are so trivial people usually don't bother to write them down. Krishna Srivastav has simply chosen to show all the steps to make it clear how the full calculation proceeds. Occasionally a good teacher has to do this, as what makes those steps trivial is that everyone understands them. However, this is not always true of new students.
– Paul Sinclair
Nov 29 at 17:29
add a comment |
1
Out of interest why can you not just go straight from the first line to the last line without needing to have the $dx/dx$ part in there too?
– Chris
Nov 29 at 17:20
1
@Chris - because going straight from the first line to the last is simply skipping over steps that are required by a direct application of the theorems, but are so trivial people usually don't bother to write them down. Krishna Srivastav has simply chosen to show all the steps to make it clear how the full calculation proceeds. Occasionally a good teacher has to do this, as what makes those steps trivial is that everyone understands them. However, this is not always true of new students.
– Paul Sinclair
Nov 29 at 17:29
1
1
Out of interest why can you not just go straight from the first line to the last line without needing to have the $dx/dx$ part in there too?
– Chris
Nov 29 at 17:20
Out of interest why can you not just go straight from the first line to the last line without needing to have the $dx/dx$ part in there too?
– Chris
Nov 29 at 17:20
1
1
@Chris - because going straight from the first line to the last is simply skipping over steps that are required by a direct application of the theorems, but are so trivial people usually don't bother to write them down. Krishna Srivastav has simply chosen to show all the steps to make it clear how the full calculation proceeds. Occasionally a good teacher has to do this, as what makes those steps trivial is that everyone understands them. However, this is not always true of new students.
– Paul Sinclair
Nov 29 at 17:29
@Chris - because going straight from the first line to the last is simply skipping over steps that are required by a direct application of the theorems, but are so trivial people usually don't bother to write them down. Krishna Srivastav has simply chosen to show all the steps to make it clear how the full calculation proceeds. Occasionally a good teacher has to do this, as what makes those steps trivial is that everyone understands them. However, this is not always true of new students.
– Paul Sinclair
Nov 29 at 17:29
add a comment |
up vote
1
down vote
Derivative of $sin(ax) = a cos(ax)$ by Chain Rule.
edited Nov 29 at 15:42
Tianlalu
2,8811935
answered Nov 29 at 15:21
Rohit Bharadwaj
518
New contributor
Rohit Bharadwaj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
1
down vote
Derivative of $sin(ax) = a cos(ax)$ by Chain Rule.
edited Nov 29 at 15:42
Tianlalu
2,8811935
answered Nov 29 at 15:21
Rohit Bharadwaj
518
New contributor
Rohit Bharadwaj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
1
down vote
up vote
1
down vote
Derivative of $sin(ax) = a cos(ax)$ by Chain Rule.
edited Nov 29 at 15:42
Tianlalu
2,8811935
answered Nov 29 at 15:21
Rohit Bharadwaj
518
New contributor
Rohit Bharadwaj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Derivative of $sin(ax) = a cos(ax)$ by Chain Rule.
edited Nov 29 at 15:42
Tianlalu
2,8811935
answered Nov 29 at 15:21
Rohit Bharadwaj
518
New contributor
Rohit Bharadwaj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Nov 29 at 15:42
Tianlalu
2,8811935
edited Nov 29 at 15:42
Tianlalu
2,8811935
edited Nov 29 at 15:42
Tianlalu
2,8811935
2,8811935
answered Nov 29 at 15:21
Rohit Bharadwaj
518
New contributor
Rohit Bharadwaj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Nov 29 at 15:21
Rohit Bharadwaj
518
answered Nov 29 at 15:21
Rohit Bharadwaj
518
518
New contributor
Rohit Bharadwaj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Rohit Bharadwaj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Rohit Bharadwaj is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
Chaku Daku is a new contributor. Be nice, and check out our Code of Conduct.
Chaku Daku is a new contributor. Be nice, and check out our Code of Conduct.
Chaku Daku is a new contributor. Be nice, and check out our Code of Conduct.
Chaku Daku is a new contributor. Be nice, and check out our Code of Conduct.
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1
Apply the chain rule.
– user3482749
Nov 29 at 15:17