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#1 2021-04-23 01:42:11
- mathland
- Member
- Registered: 2021-03-25
- Posts: 444
Product Rule
Given g(s) = 2s/(s + 1), find g'(s).
Let me see.
g'(s) = [(s + 1)d/ds (2s) - (2s)d/ds (s + 1)]/(s + 1)^2
g'(s) = [(s + 1)(2) - (2s)(1)]/(s + 1)^2
g'(s) = (2s + 2 - 2s)/(s + 1)^2
g'(s) = 2/(s + 1)^2
Correct?
What I found is the slope of the given curve at any point.
Correct?
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#2 2021-04-23 02:12:45
- Mathegocart
- Member
- Registered: 2012-04-29
- Posts: 2,226
Re: Product Rule
Given g(s) = 2s/(s + 1), find g'(s).
Let me see.
g'(s) = [(s + 1)d/ds (2s) - (2s)d/ds (s + 1)]/(s + 1)^2
g'(s) = [(s + 1)(2) - (2s)(1)]/(s + 1)^2
g'(s) = (2s + 2 - 2s)/(s + 1)^2
g'(s) = 2/(s + 1)^2
Correct?
What I found is the slope of the given curve at any point.
Correct?
Your answer and your working is all OK. Wonderful work.
The derivative is the slope of the tangent line. So yes, that is the slope of the given curve at any point.
The integral of hope is reality.
May bobbym have a wonderful time in the pearly gates of heaven.
He will be sorely missed.
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#3 2021-04-23 10:39:03
- mathland
- Member
- Registered: 2021-03-25
- Posts: 444
Re: Product Rule
mathland wrote:Given g(s) = 2s/(s + 1), find g'(s).
Let me see.
g'(s) = [(s + 1)d/ds (2s) - (2s)d/ds (s + 1)]/(s + 1)^2
g'(s) = [(s + 1)(2) - (2s)(1)]/(s + 1)^2
g'(s) = (2s + 2 - 2s)/(s + 1)^2
g'(s) = 2/(s + 1)^2
Correct?
What I found is the slope of the given curve at any point.
Correct?
Your answer and your working is all OK. Wonderful work.
The derivative is the slope of the tangent line. So yes, that is the slope of the given curve at any point.
It's wonderful to know I got it right.
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