Question

Derivative help.

Answer 1

d/dx (cosx)/e^x)

Answer 2

http://www.wolframalpha.com/input/?i=d%2Fdx+%28cosx%29%2Fe%5Ex%29

this should give your answer.

Answer 3

Dx(cos(x)*e^(-x)) and product rule it

Answer 4

yeah, either use the product rule or the quotient rule (which in fact are equivalent)

Answer 5

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.

Answer 6

d/dx((cos(x))/e^x) = -e^(-x) (sin(x)+cos(x))

Answer 7

cos'(x)*e^(-x) + cos(x)*e'^(-x)

-sin(x)*e^(-x) - cos(x)*e^(-x)

Answer 8

\[\frac{d}{dx} \left(\frac{\cos (x)}{e^x}\right) = \frac{d}{dx} e^{-x}\cos (x) = e^{-x}(-\sin (x)) + \cos (x)(-e^{-x}) = -e^{-x} (\sin (x) + \cos (x))\]

Answer 9

hmm it wrote it off the page

Answer 10

\[\frac{d}{dx} \left(\frac{\cos (x)}{e^x}\right) = \frac{d}{dx} e^{-x}\cos (x)\]
\[ = e^{-x}(-\sin (x)) + \cos (x)(-e^{-x})\]
\[ = -e^{-x} (\sin (x) + \cos (x))\]

Answer 11

In the second step why the second term is e^-x, why you did not derviate it ?

Answer 12

the derivative of \[e^{-x}\] is \[-e^{-x}\]

Answer 13

but i learned the derivative of e^x is e^x

Answer 14

this is what khan academy says the answer is

Answer 15

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

\[\frac{dy}{dx} = \frac{dy}{dt}\frac{dt}{dx} = e^{t}(-1) = e^{-x}(-1) = -e^{-x}\]

Answer 16

Yes, notice that answer a) is exactly the same as \[-e^{-x} (\sin (x) + \cos (x))\]

Answer 17

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.

Answer 18

Thanks a lot, i understood it.