Question

Determine the derivative of each of the following:...
f(x)= e^-x^3/x

Answer 1

@Jamierox4ev3r

Answer 2

so that be f1(x)= e^-x^3 . -x

Answer 3

This is the function?\[\Large\rm f(x)= \frac{e^{-x^3}}{x}\]

Answer 4

yes

Answer 5

So you'll have to apply quotient rule, ya?

Answer 6

Here is our quotient rule setup,\[\Large\rm f'(x)= \frac{\color{royalblue}{\left(e^{-x^3}\right)'}x-e^{-x^3}\color{royalblue}{\left(x\right)'}}{x^2}\]Derivatives need to be taken for the blue parts.

Answer 7

f1(x)= -3x^2 e (x) - e^-x^3(1)/ (x)^2

Answer 8

right??

Answer 9

@MTALHAHASSAN2 you took the derivative of `e^(-x^3)` incorrectly

Answer 10

Hmm, I'm just a little bit worried about your first e,
did you forget to write the exponent?

Answer 11

f1(x)= e^ -3x^2

Answer 12

Is it be like that then

Answer 13

Recall: \(\large\rm \dfrac{d}{dx}e^{x}=e^x\)
You get the same exponential back.
We need to simply apply the chain rule as an added step.
\[\large\rm \frac{d}{dx}e^{-x^3}\quad=e^{-x^3}\frac{d}{dx}(-x^3)\quad=e^{-x^3}(-3x^2)\]

Answer 14

You get the same exponential back (unchanged),
plus some other stuff, due to the chain rule.

Answer 15

oh ok

Answer 16

times* some other stuff, due to the chain rule.

that word plus is a lil misleading :) lol

Answer 17

\[\Large\rm f'(x)= \frac{\color{royalblue}{\left(e^{-x^3}\right)'}x-e^{-x^3}\color{royalblue}{\left(x\right)'}}{x^2}\]
\[\Large\rm f'(x)= \frac{\color{orangered}{\left(-3x^2e^{-x^3}\right)}x-e^{-x^3}\color{orangered}{\left(1\right)}}{x^2}\]

Answer 18

And then simplify some stuff, ya? :)

Answer 19

yeah

Answer 20

are we simplify the x right

Answer 21

f1(x)= -3x^3 e^-x^4

Answer 22

Hmmm, I'm not sure what you did...

Combine the x's in the first term,\[\Large\rm f'(x)= \frac{-3x^3e^{-x^3}-e^{-x^3}}{x^2}\]Then factor a -e^(-x^3) out of each term,\[\Large\rm f'(x)= \frac{-e^{-x^3}\left(~?~~+~~?~\right)}{x^2}\]What are you left with in the brackets?
Should be two terms still.

Answer 23

(-3x^2+1)

Answer 24

right??

Answer 25

@zepdrix

Answer 26

Hmm, both terms were negative,
so when you pull that negative out,
they both become positive inside of the brackets, ya?

Answer 27

(3x^2+1)

Looks good besides that though :)

Answer 28

Woops*
(3x^3+1)

Answer 29

but at the book of the book they have a different answer