## Jusaquikie Group Title Find the derivative of the function. y = 5xe^−kx y'(x) = 2 years ago 2 years ago

1. Jusaquikie

$y = 5xe^{-kx}$

2. Jusaquikie

i know to use the chain rule and the product rule but i don't know how to seperate this out properly

3. KKJ

y is made up of a product function. Thus, 5x and e^(-kx) Hence, we are to use the product rule with the above separations

4. Jusaquikie

the answer has ln in it how do e and ln work together?

5. KKJ

Where from the In?

6. KKJ

If you differentiate y = e^ax, what do you get?

7. Jusaquikie

it would be E^ax right?

8. Jusaquikie

or would it be ae^ax-1

9. KKJ

$\frac{ dy }{ dx } = u \frac{ dv }{ dx } + v\frac{ du }{ dx }$

10. Jusaquikie

or (e^ax)a

11. KKJ

it will give you ae^ax

12. Jusaquikie

constant letters confuse me more than anything

13. KKJ

Consider them as numbers

14. Jusaquikie

but moving a to the front of the equason wouldn't i have a -1 on the exponent?

15. KKJ

no

16. Jusaquikie

so 5xe^-kx is 5x*e^-kx so (5)(e^-kx)+(5x)(-ke^-kx) = 5e^-kx - 5xke^-kx

17. KKJ

So $y=5xe^\left( -kx \right)$ $\frac{ dy }{ dx } = 5x \frac{ d }{ dx }(e^{-kx}) + e ^{-kx} \frac{ d }{ dx }(5x)$

18. KKJ

$\frac{ dy }{ dx } = 5x(-ke ^{-kx}) + e ^{-kx}(5)$

19. KKJ

$\frac{ dy }{ dx } = -5kxe ^{-kx} + 5e ^{-kx}$

20. KKJ

$\frac{ dy }{ dx } = -5e ^{-kx} (kx - 1)$

21. Jusaquikie

Thanks KKJ