Calculating the derivative of $sin^3(312x^2)$ [closed]
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Can you explain in detail please how to find the derivative of this function?
$$sin^3(312x^2)$$
calculus derivatives
closed as off-topic by Ennar, A. Goodier, José Carlos Santos, Shailesh, Leucippus Nov 18 at 1:03
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." – Ennar, A. Goodier, José Carlos Santos, Shailesh, Leucippus
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Can you explain in detail please how to find the derivative of this function?
$$sin^3(312x^2)$$
calculus derivatives
closed as off-topic by Ennar, A. Goodier, José Carlos Santos, Shailesh, Leucippus Nov 18 at 1:03
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." – Ennar, A. Goodier, José Carlos Santos, Shailesh, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
1
Hi and welcome to MSE. Questions are mostly well received after the OP has showed some effort or thoughts. Did you have any attempts on the specific differentiation ?
– Rebellos
Nov 17 at 15:49
What have you tried? Do you know the power rule and the chain rule for derivatives?
– Dave
Nov 17 at 15:49
What is the derivative of $sin^3(x)$?
– Robert Z
Nov 17 at 15:50
How to ask a good question.
– Ennar
Nov 17 at 15:50
Try the Chain rule
– Sauhard Sharma
Nov 17 at 15:53
|
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Can you explain in detail please how to find the derivative of this function?
$$sin^3(312x^2)$$
calculus derivatives
Can you explain in detail please how to find the derivative of this function?
$$sin^3(312x^2)$$
calculus derivatives
calculus derivatives
edited Nov 17 at 15:48
Rebellos
12.9k21042
12.9k21042
asked Nov 17 at 15:45
Александр
1
1
closed as off-topic by Ennar, A. Goodier, José Carlos Santos, Shailesh, Leucippus Nov 18 at 1:03
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." – Ennar, A. Goodier, José Carlos Santos, Shailesh, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Ennar, A. Goodier, José Carlos Santos, Shailesh, Leucippus Nov 18 at 1:03
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." – Ennar, A. Goodier, José Carlos Santos, Shailesh, Leucippus
If this question can be reworded to fit the rules in the help center, please edit the question.
1
Hi and welcome to MSE. Questions are mostly well received after the OP has showed some effort or thoughts. Did you have any attempts on the specific differentiation ?
– Rebellos
Nov 17 at 15:49
What have you tried? Do you know the power rule and the chain rule for derivatives?
– Dave
Nov 17 at 15:49
What is the derivative of $sin^3(x)$?
– Robert Z
Nov 17 at 15:50
How to ask a good question.
– Ennar
Nov 17 at 15:50
Try the Chain rule
– Sauhard Sharma
Nov 17 at 15:53
|
show 1 more comment
1
Hi and welcome to MSE. Questions are mostly well received after the OP has showed some effort or thoughts. Did you have any attempts on the specific differentiation ?
– Rebellos
Nov 17 at 15:49
What have you tried? Do you know the power rule and the chain rule for derivatives?
– Dave
Nov 17 at 15:49
What is the derivative of $sin^3(x)$?
– Robert Z
Nov 17 at 15:50
How to ask a good question.
– Ennar
Nov 17 at 15:50
Try the Chain rule
– Sauhard Sharma
Nov 17 at 15:53
1
1
Hi and welcome to MSE. Questions are mostly well received after the OP has showed some effort or thoughts. Did you have any attempts on the specific differentiation ?
– Rebellos
Nov 17 at 15:49
Hi and welcome to MSE. Questions are mostly well received after the OP has showed some effort or thoughts. Did you have any attempts on the specific differentiation ?
– Rebellos
Nov 17 at 15:49
What have you tried? Do you know the power rule and the chain rule for derivatives?
– Dave
Nov 17 at 15:49
What have you tried? Do you know the power rule and the chain rule for derivatives?
– Dave
Nov 17 at 15:49
What is the derivative of $sin^3(x)$?
– Robert Z
Nov 17 at 15:50
What is the derivative of $sin^3(x)$?
– Robert Z
Nov 17 at 15:50
How to ask a good question.
– Ennar
Nov 17 at 15:50
How to ask a good question.
– Ennar
Nov 17 at 15:50
Try the Chain rule
– Sauhard Sharma
Nov 17 at 15:53
Try the Chain rule
– Sauhard Sharma
Nov 17 at 15:53
|
show 1 more comment
3 Answers
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The derivative of $f(g(x))$ is (if it exists) $$f'(g(x))g'(x)tag1$$ (application of chain rule).
Here $g(x)=312x^2$ (can you find $g'(x)$?).
And $f(x)=sin^3(x)$.
Note that we can write $f(x)=h(k(x))$ where $h(x)=x^3$ and $k(x)=sin x$ so that according to $(1)$: $$f'(x)=h'(k(x))k'(x)$$
This must be enough.
add a comment |
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This is a repeated application of the chain rule:
$$frac{d}{dx} sin^3(312x^2) = 3 sin^2(312x^2) cdot frac{d}{dx} (sin(312x^2)= 3 sin^2(312x^2) cdot cos(312x^2) cdot frac{d}{dx} 312x^2 =\ 3 sin^2(312x^2) cdot cos(312x^2) cdot 624x$$
add a comment |
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$$y = sin^3(312x^2)$$
This is a composite function. Whenever you have a composite function, use the Chain Rule.
$$(fcirc g)’(x) = f’(g(x))cdot g’x$$
Let $f(x) = x^3$ and $g(x) = sin(312x^2)$.
$$implies y’ = 3sin^2(312x^2)cdot [sin (312x^2)]’$$
Here, you have to apply the same technique again to simplify the second part.
$$[sin(312x^2)]’ = sin’(312x^2)cdot (312x^2)’ = cos(312x^2)cdot 64x = 624xcos(312x^2)$$
So, the expression becomes
$$y’ = 3sin^2(312x^2)cdot 624xcos(312x^2) = 1872xsin^2(312x^2)cos(312x^2)$$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
The derivative of $f(g(x))$ is (if it exists) $$f'(g(x))g'(x)tag1$$ (application of chain rule).
Here $g(x)=312x^2$ (can you find $g'(x)$?).
And $f(x)=sin^3(x)$.
Note that we can write $f(x)=h(k(x))$ where $h(x)=x^3$ and $k(x)=sin x$ so that according to $(1)$: $$f'(x)=h'(k(x))k'(x)$$
This must be enough.
add a comment |
up vote
1
down vote
The derivative of $f(g(x))$ is (if it exists) $$f'(g(x))g'(x)tag1$$ (application of chain rule).
Here $g(x)=312x^2$ (can you find $g'(x)$?).
And $f(x)=sin^3(x)$.
Note that we can write $f(x)=h(k(x))$ where $h(x)=x^3$ and $k(x)=sin x$ so that according to $(1)$: $$f'(x)=h'(k(x))k'(x)$$
This must be enough.
add a comment |
up vote
1
down vote
up vote
1
down vote
The derivative of $f(g(x))$ is (if it exists) $$f'(g(x))g'(x)tag1$$ (application of chain rule).
Here $g(x)=312x^2$ (can you find $g'(x)$?).
And $f(x)=sin^3(x)$.
Note that we can write $f(x)=h(k(x))$ where $h(x)=x^3$ and $k(x)=sin x$ so that according to $(1)$: $$f'(x)=h'(k(x))k'(x)$$
This must be enough.
The derivative of $f(g(x))$ is (if it exists) $$f'(g(x))g'(x)tag1$$ (application of chain rule).
Here $g(x)=312x^2$ (can you find $g'(x)$?).
And $f(x)=sin^3(x)$.
Note that we can write $f(x)=h(k(x))$ where $h(x)=x^3$ and $k(x)=sin x$ so that according to $(1)$: $$f'(x)=h'(k(x))k'(x)$$
This must be enough.
edited Nov 17 at 16:08
Ethan Bolker
40.1k544106
40.1k544106
answered Nov 17 at 15:56
drhab
95.1k543126
95.1k543126
add a comment |
add a comment |
up vote
0
down vote
This is a repeated application of the chain rule:
$$frac{d}{dx} sin^3(312x^2) = 3 sin^2(312x^2) cdot frac{d}{dx} (sin(312x^2)= 3 sin^2(312x^2) cdot cos(312x^2) cdot frac{d}{dx} 312x^2 =\ 3 sin^2(312x^2) cdot cos(312x^2) cdot 624x$$
add a comment |
up vote
0
down vote
This is a repeated application of the chain rule:
$$frac{d}{dx} sin^3(312x^2) = 3 sin^2(312x^2) cdot frac{d}{dx} (sin(312x^2)= 3 sin^2(312x^2) cdot cos(312x^2) cdot frac{d}{dx} 312x^2 =\ 3 sin^2(312x^2) cdot cos(312x^2) cdot 624x$$
add a comment |
up vote
0
down vote
up vote
0
down vote
This is a repeated application of the chain rule:
$$frac{d}{dx} sin^3(312x^2) = 3 sin^2(312x^2) cdot frac{d}{dx} (sin(312x^2)= 3 sin^2(312x^2) cdot cos(312x^2) cdot frac{d}{dx} 312x^2 =\ 3 sin^2(312x^2) cdot cos(312x^2) cdot 624x$$
This is a repeated application of the chain rule:
$$frac{d}{dx} sin^3(312x^2) = 3 sin^2(312x^2) cdot frac{d}{dx} (sin(312x^2)= 3 sin^2(312x^2) cdot cos(312x^2) cdot frac{d}{dx} 312x^2 =\ 3 sin^2(312x^2) cdot cos(312x^2) cdot 624x$$
answered Nov 17 at 16:06
Henno Brandsma
102k345109
102k345109
add a comment |
add a comment |
up vote
0
down vote
$$y = sin^3(312x^2)$$
This is a composite function. Whenever you have a composite function, use the Chain Rule.
$$(fcirc g)’(x) = f’(g(x))cdot g’x$$
Let $f(x) = x^3$ and $g(x) = sin(312x^2)$.
$$implies y’ = 3sin^2(312x^2)cdot [sin (312x^2)]’$$
Here, you have to apply the same technique again to simplify the second part.
$$[sin(312x^2)]’ = sin’(312x^2)cdot (312x^2)’ = cos(312x^2)cdot 64x = 624xcos(312x^2)$$
So, the expression becomes
$$y’ = 3sin^2(312x^2)cdot 624xcos(312x^2) = 1872xsin^2(312x^2)cos(312x^2)$$
add a comment |
up vote
0
down vote
$$y = sin^3(312x^2)$$
This is a composite function. Whenever you have a composite function, use the Chain Rule.
$$(fcirc g)’(x) = f’(g(x))cdot g’x$$
Let $f(x) = x^3$ and $g(x) = sin(312x^2)$.
$$implies y’ = 3sin^2(312x^2)cdot [sin (312x^2)]’$$
Here, you have to apply the same technique again to simplify the second part.
$$[sin(312x^2)]’ = sin’(312x^2)cdot (312x^2)’ = cos(312x^2)cdot 64x = 624xcos(312x^2)$$
So, the expression becomes
$$y’ = 3sin^2(312x^2)cdot 624xcos(312x^2) = 1872xsin^2(312x^2)cos(312x^2)$$
add a comment |
up vote
0
down vote
up vote
0
down vote
$$y = sin^3(312x^2)$$
This is a composite function. Whenever you have a composite function, use the Chain Rule.
$$(fcirc g)’(x) = f’(g(x))cdot g’x$$
Let $f(x) = x^3$ and $g(x) = sin(312x^2)$.
$$implies y’ = 3sin^2(312x^2)cdot [sin (312x^2)]’$$
Here, you have to apply the same technique again to simplify the second part.
$$[sin(312x^2)]’ = sin’(312x^2)cdot (312x^2)’ = cos(312x^2)cdot 64x = 624xcos(312x^2)$$
So, the expression becomes
$$y’ = 3sin^2(312x^2)cdot 624xcos(312x^2) = 1872xsin^2(312x^2)cos(312x^2)$$
$$y = sin^3(312x^2)$$
This is a composite function. Whenever you have a composite function, use the Chain Rule.
$$(fcirc g)’(x) = f’(g(x))cdot g’x$$
Let $f(x) = x^3$ and $g(x) = sin(312x^2)$.
$$implies y’ = 3sin^2(312x^2)cdot [sin (312x^2)]’$$
Here, you have to apply the same technique again to simplify the second part.
$$[sin(312x^2)]’ = sin’(312x^2)cdot (312x^2)’ = cos(312x^2)cdot 64x = 624xcos(312x^2)$$
So, the expression becomes
$$y’ = 3sin^2(312x^2)cdot 624xcos(312x^2) = 1872xsin^2(312x^2)cos(312x^2)$$
answered Nov 17 at 16:19
KM101
3,316417
3,316417
add a comment |
add a comment |
1
Hi and welcome to MSE. Questions are mostly well received after the OP has showed some effort or thoughts. Did you have any attempts on the specific differentiation ?
– Rebellos
Nov 17 at 15:49
What have you tried? Do you know the power rule and the chain rule for derivatives?
– Dave
Nov 17 at 15:49
What is the derivative of $sin^3(x)$?
– Robert Z
Nov 17 at 15:50
How to ask a good question.
– Ennar
Nov 17 at 15:50
Try the Chain rule
– Sauhard Sharma
Nov 17 at 15:53