## anonymous 4 years ago Derivative help.

1. anonymous

d/dx (cosx)/e^x)

2. anonymous

3. amistre64

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

4. anonymous

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

5. 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.

6. anonymous

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

7. amistre64

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

8. anonymous

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

9. anonymous

hmm it wrote it off the page

10. anonymous

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

11. anonymous

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

12. anonymous

the derivative of $e^{-x}$ is $-e^{-x}$

13. anonymous

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

14. anonymous

this is what khan academy says the answer is

15. anonymous

16. 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 $\frac{dy}{dx} = \frac{dy}{dt}\frac{dt}{dx} = e^{t}(-1) = e^{-x}(-1) = -e^{-x}$

17. anonymous

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

18. 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.

19. anonymous

Thanks a lot, i understood it.