anonymous
  • anonymous
Derivative help.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
d/dx (cosx)/e^x)
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=d%2Fdx+%28cosx%29%2Fe%5Ex%29 this should give your answer.
amistre64
  • amistre64
Dx(cos(x)*e^(-x)) and product rule it

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anonymous
  • anonymous
yeah, either use the product rule or the quotient rule (which in fact are equivalent)
anonymous
  • anonymous
I got the answer but it's different what khanAcademy is showing me, so please give me the exact answer so that i can verify it.
anonymous
  • anonymous
d/dx((cos(x))/e^x) = -e^(-x) (sin(x)+cos(x))
amistre64
  • amistre64
cos'(x)*e^(-x) + cos(x)*e'^(-x) -sin(x)*e^(-x) - cos(x)*e^(-x)
anonymous
  • anonymous
\[\frac{d}{dx} \left(\frac{\cos (x)}{e^x}\right) = \frac{d}{dx} e^{-x}\cos (x) = e^{-x}(-\sin (x)) + \cos (x)(-e^{-x}) = -e^{-x} (\sin (x) + \cos (x))\]
anonymous
  • anonymous
hmm it wrote it off the page
anonymous
  • anonymous
\[\frac{d}{dx} \left(\frac{\cos (x)}{e^x}\right) = \frac{d}{dx} e^{-x}\cos (x)\] \[ = e^{-x}(-\sin (x)) + \cos (x)(-e^{-x})\] \[ = -e^{-x} (\sin (x) + \cos (x))\]
anonymous
  • anonymous
In the second step why the second term is e^-x, why you did not derviate it ?
anonymous
  • anonymous
the derivative of \[e^{-x}\] is \[-e^{-x}\]
anonymous
  • anonymous
but i learned the derivative of e^x is e^x
anonymous
  • anonymous
this is what khan academy says the answer is
anonymous
  • anonymous
anonymous
  • anonymous
yes that's right. Let's show, usin gthe chain rule that the derivative of e^{-x} is -e^{-x}. Find the derivative of \[y = e^{-x}. \] Let t=-x. Then y = e^t \[\frac{dy}{dx} = \frac{dy}{dt}\frac{dt}{dx} = e^{t}(-1) = e^{-x}(-1) = -e^{-x}\]
anonymous
  • anonymous
Yes, notice that answer a) is exactly the same as \[-e^{-x} (\sin (x) + \cos (x))\]
anonymous
  • anonymous
That answer is likely derived from the quotient rule, which can be re-written as the product rule by making the requirement of multiplying by e^{-x} rather than dividing by e^x. Notice that this is equivalent.
anonymous
  • anonymous
Thanks a lot, i understood it.

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