What is derivative of $sin ax$ where $a$ is a constant?
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 '18 at 19:19
Asaf Karagila♦
302k32427757
asked Nov 29 '18 at 15:17
Chaku DakuChaku Daku
91
add a comment |
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 '18 at 19:19
Asaf Karagila♦
302k32427757
asked Nov 29 '18 at 15:17
Chaku DakuChaku Daku
91
1
Apply the chain rule.
– user3482749
Nov 29 '18 at 15:17
add a comment |
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 '18 at 19:19
Asaf Karagila♦
302k32427757
asked Nov 29 '18 at 15:17
Chaku DakuChaku Daku
91
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 '18 at 19:19
Asaf Karagila♦
302k32427757
asked Nov 29 '18 at 15:17
Chaku DakuChaku Daku
91
edited Nov 29 '18 at 19:19
Asaf Karagila♦
302k32427757
asked Nov 29 '18 at 15:17
Chaku DakuChaku Daku
91
edited Nov 29 '18 at 19:19
Asaf Karagila♦
302k32427757
edited Nov 29 '18 at 19:19
Asaf Karagila♦
302k32427757
edited Nov 29 '18 at 19:19
Asaf Karagila♦
302k32427757
302k32427757
asked Nov 29 '18 at 15:17
Chaku DakuChaku Daku
91
asked Nov 29 '18 at 15:17
Chaku DakuChaku Daku
91
asked Nov 29 '18 at 15:17
Chaku DakuChaku Daku
91
91
1
Apply the chain rule.
– user3482749
Nov 29 '18 at 15:17
add a comment |
1
Apply the chain rule.
– user3482749
Nov 29 '18 at 15:17
1
1
Apply the chain rule.
– user3482749
Nov 29 '18 at 15:17
Apply the chain rule.
– user3482749
Nov 29 '18 at 15:17
add a comment |
3 Answers
3
active
oldest
votes
HINT
Recall that by chain rule
$$frac{d}{dx}[sin (f(x))]=cos (f(x))cdot f'(x)$$
answered Nov 29 '18 at 15:20
gimusigimusi
1
add a comment |
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 '18 at 16:10
Krishna SrivastavKrishna 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 '18 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 '18 at 17:29
add a comment |
Derivative of $sin(ax) = a cos(ax)$ by Chain Rule.
edited Nov 29 '18 at 15:42
Tianlalu
3,09621038
answered Nov 29 '18 at 15:21
Rohit BharadwajRohit Bharadwaj
518
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
HINT
Recall that by chain rule
$$frac{d}{dx}[sin (f(x))]=cos (f(x))cdot f'(x)$$
answered Nov 29 '18 at 15:20
gimusigimusi
1
add a comment |
HINT
Recall that by chain rule
$$frac{d}{dx}[sin (f(x))]=cos (f(x))cdot f'(x)$$
answered Nov 29 '18 at 15:20
gimusigimusi
1
add a comment |
HINT
Recall that by chain rule
$$frac{d}{dx}[sin (f(x))]=cos (f(x))cdot f'(x)$$
answered Nov 29 '18 at 15:20
gimusigimusi
1
HINT
Recall that by chain rule
$$frac{d}{dx}[sin (f(x))]=cos (f(x))cdot f'(x)$$
answered Nov 29 '18 at 15:20
gimusigimusi
1
answered Nov 29 '18 at 15:20
gimusigimusi
1
answered Nov 29 '18 at 15:20
gimusigimusi
1
answered Nov 29 '18 at 15:20
gimusigimusi
1
1
add a comment |
add a comment |
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 '18 at 16:10
Krishna SrivastavKrishna 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 '18 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 '18 at 17:29
add a comment |
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 '18 at 16:10
Krishna SrivastavKrishna 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 '18 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 '18 at 17:29
add a comment |
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 '18 at 16:10
Krishna SrivastavKrishna 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 '18 at 16:10
Krishna SrivastavKrishna Srivastav
894
answered Nov 29 '18 at 16:10
Krishna SrivastavKrishna Srivastav
894
answered Nov 29 '18 at 16:10
Krishna SrivastavKrishna Srivastav
894
answered Nov 29 '18 at 16:10
Krishna SrivastavKrishna 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 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 '18 at 17:29
add a comment |
Derivative of $sin(ax) = a cos(ax)$ by Chain Rule.
edited Nov 29 '18 at 15:42
Tianlalu
3,09621038
answered Nov 29 '18 at 15:21
Rohit BharadwajRohit Bharadwaj
518
add a comment |
Derivative of $sin(ax) = a cos(ax)$ by Chain Rule.
edited Nov 29 '18 at 15:42
Tianlalu
3,09621038
answered Nov 29 '18 at 15:21
Rohit BharadwajRohit Bharadwaj
518
add a comment |
Derivative of $sin(ax) = a cos(ax)$ by Chain Rule.
edited Nov 29 '18 at 15:42
Tianlalu
3,09621038
answered Nov 29 '18 at 15:21
Rohit BharadwajRohit Bharadwaj
518
Derivative of $sin(ax) = a cos(ax)$ by Chain Rule.
edited Nov 29 '18 at 15:42
Tianlalu
3,09621038
answered Nov 29 '18 at 15:21
Rohit BharadwajRohit Bharadwaj
518
edited Nov 29 '18 at 15:42
Tianlalu
3,09621038
edited Nov 29 '18 at 15:42
Tianlalu
3,09621038
edited Nov 29 '18 at 15:42
Tianlalu
3,09621038
3,09621038
answered Nov 29 '18 at 15:21
Rohit BharadwajRohit Bharadwaj
518
answered Nov 29 '18 at 15:21
Rohit BharadwajRohit Bharadwaj
518
answered Nov 29 '18 at 15:21
Rohit BharadwajRohit Bharadwaj
518
518
add a comment |
add a comment |
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1
Apply the chain rule.
– user3482749
Nov 29 '18 at 15:17