Question 1002128: differentiate the following
f(s)=(1+s+e^s)(2e^-s+s)
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! f(s)=(1+s+e^s)(2e^-s+s)
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d/ds((1+s+e^s) (2/e^s+s))
Rewrite the expression: (1+s+e^s) (2/e^s+s) = (2/e^s+s) (1+e^s+s):
= d/ds((2/e^s+s) (1+e^s+s))
Use the product rule, d/ds(u v) = v ( du)/( ds) + u ( dv)/( ds), where u = s+2 e^(-s) and v = s+e^s+1:
= (1+e^s+s) (d/ds(2/e^s+s))+(2 e^(-s)+s) (d/ds(1+e^s+s))
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Differentiate the sum term by term and factor out constants:
= (2/e^s+s) (d/ds(1+e^s+s))+(1+e^s+s) 2 d/ds(e^(-s))+d/ds(s)
Using the chain rule, d/ds(e^(-s)) = ( de^u)/( du) ( du)/( ds), where u = -s and ( d)/( du)(e^u) = e^u:
= (2/e^s+s) (d/ds(1+e^s+s))+(1+e^s+s) (d/ds(s)+2 (d/ds(-s))/e^s)
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Factor out constants:
= (2/e^s+s) (d/ds(1+e^s+s))+(1+e^s+s) (d/ds(s)+(2 -d/ds(s))/e^s)
Simplify the expression:
= (1+e^s+s) (d/ds(s)-(2 (d/ds(s)))/e^s)+(2/e^s+s) (d/ds(1+e^s+s))
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The derivative of s is 1:
= (2/e^s+s) (d/ds(1+e^s+s))+(1+e^s+s) (d/ds(s)-(1 2)/e^s)
The derivative of s is 1:
= (2/e^s+s) (d/ds(1+e^s+s))+(1+e^s+s) (-2/e^s+1)
Differentiate the sum term by term:
= (1-2/e^s) (1+e^s+s)+(2/e^s+s) d/ds(1)+d/ds(e^s)+d/ds(s)
The derivative of 1 is zero:
= (1-2/e^s) (1+e^s+s)+(2/e^s+s) (d/ds(e^s)+d/ds(s)+0)
Simplify the expression:
= (1-2/e^s) (1+e^s+s)+(2/e^s+s) (d/ds(e^s)+d/ds(s))
The derivative of e^s is e^s:
= (1-2/e^s) (1+e^s+s)+(2/e^s+s) (d/ds(s)+e^s)
The derivative of s is 1:
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Answer: = (1-2/e^s) (1+e^s+s)+(2/e^s+s) (e^s+1)
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