Question

Find the derivative of the function.
y = 5xe^−kx
y'(x) =

Answer 1

\[ y = 5xe^{-kx}\]

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

Where from the In?

Answer 6

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

Answer 7

it would be E^ax right?

Answer 8

or would it be ae^ax-1

Answer 9

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

Answer 10

or (e^ax)a

Answer 11

it will give you ae^ax

Answer 12

constant letters confuse me more than anything

Answer 13

Consider them as numbers

Answer 14

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

Answer 15

no

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

Thanks KKJ