Question

x^2e^(-ax)
How do I differentiate ?

Answer 1

split it into smaller functions that are easy to differentiate

x^2
e^(-ax)

then use product rule

Answer 2

thank god, someone is here
how do I know to use the chain rule or product rule first?

Answer 3

how do I use product rule when there is that (-ax) part

Answer 4

Always split the function into smaller functions take the derivative then apply the rules

Answer 5

The derivative of
e^u is (u')e^u

Answer 6

the derivative of

e^x is (1)e^x = e^(x)

Answer 7

how does that work into the product rule considering there is only f g f' g'

Answer 8

First make up two new functions out of this function

so

g(x) = x^2

s(x) = e^(-ax)

find the derivative of both of these functions then just use product rule (MEMORIZE THIS)

f'(x) = g'(x)s(x) + s'(x)g(x)

Answer 9

Their is really no trick to derivatives just split your function into the smallest functions possible then just apply rules

Answer 10

you really only need to memorize a few chain, product rule, quotient rule (although you can just use product rule instead)

Answer 11

I know the product rule I guess my question is this
if I had X^2e^x I would just use product rule cause its two things X^2 and e^x, one is f the other is g
But with e^-ax there is also the derivative of -ax, where does that fit in, into the product rule?

Answer 12

is that

\[e^{-ax}\]
or
\[e^{-x} a\]

Answer 13

the first one

Answer 14

have you covered partial derivatives in class?

Answer 15

? alright

Answer 16

Yes I think so sorry I was looking at it here

Answer 17

im not sure what a partial derivative is though

Answer 18

well I would just treat it as a constant

Answer 19

Do you know how to compute the derivative or do you still need help?

Answer 20

oh boy yea I would like to know once and for all what the derivative is :)

Answer 21

well split it into smaller functions

g(x) = x^2

s(x) = e^(-ax)

finding the derivative of e^(-ax), split it further

m(x) = e^(x)

d(x) = -ax

apply (d'(x))m(dx)


then just put it all together and apply product rule

Answer 22

d(x) = -ax
d'(x) = -a(1) = -a

Answer 23

s(x) = e^(-ax)
s'(x) = (-a)e^(-ax)

g(x) = x^2
g'(x) = 2x

f'(x) = g(x)s'(x) + g'(x)s(x)

Answer 24

ok so s'(x) is both the derivative of (-ax) and e^(-ax)
this is what I was not sure of

Answer 25

yeah

Answer 26

the derivative of

e^f(x)
is

f'(x)e^f(x)

Answer 27

yes I am familiar with that

Answer 28

so e^(-ax)

assuming, "a" is a constant would just be the same

Answer 29

if it was a third variable you would have to use partial derivatives

Answer 30

the answer is 0, and 2/a

Answer 31

f'(x) = 2x(e^(-ax)) + x^2(-ae^(-ax))

Answer 32

im trying to figure that out now
so I get
2x(e^(-ax)+x^2*-ae^(-ax)

Answer 33

so the given answer is 0 or 2/a?

Answer 34

sweeet

Answer 35

x= 0, 2/a
so both

Answer 36

Are you looking to make f'(x) = 0

Answer 37

find a formula for the x coordinates fo the critical points of f in terms of a

Answer 38

alright fair enough

Answer 39

I'm thinking if a is your y value then you need to use implicit differentiation

Answer 40

how do you simplify the equation we got Im having a hard time getting it down to 2/a

Answer 41

critical points are where your f'(x) = 0

Answer 42

no this is just introductory derivatives from calc one before L Hopitals rule and everything
this is just after learning chain rule, product rule

Answer 43

ok, but yeah critical points are found by setting f'(x) = 0 and also by checking the domain of f'(x)

Answer 44

critical points are an indication where the f(x) changes

Answer 45

yea I reduce and then set to zero

Answer 46

yea crit points are f'(x)
inflection points are f''(x) (don't need that for this equation though)

Answer 47

so you factor out e^(-ax)?

Answer 48

0 = 2x(e^(-ax)+x^2*-ae^(-ax)

so one of the critical points will be 0

0 = 2(0)(e^(-ax)+(0)^2*-ae^(-ax)
0 = 0

I dont see any other critical points possible

Answer 49

https://www.wolframalpha.com/input/?i=+0+%3D+2x%28e^%28-ax%29+-+x^2*ae^%28-ax%29

Answer 50

So you haven't seen implicit differentiation?

Answer 51

nope we don't deal with 3 variables yet

Answer 52

implicit differentiation isn't dealing with three variables

Answer 53

we are taking the derivative da/dx I think

Answer 54

well in this example I have to algebraically simplify then solve both sides to 0

Answer 55

0 = xe^(-ax)(2-ax)

Answer 56

On implicit differentiation, if it is then we have to do this derivative differently
https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/implicitdiffdirectory/ImplicitDiff.html

Answer 57

ah ha! yes that last equation is it!

Answer 58

oh I guess a is a constant

Answer 59

I'm just tired ha

Answer 60

yeah I just factored it

Answer 61

I set xe^(-ax) to 0 and get 0 then I solve for x in 2-ax to get 2/a?

Answer 62

set
(2-ax) = 0
set
xe^(-ax) = 0

and solve for x

Answer 63

remember e^(x) =/= 0

Answer 64

I have a question can 2-ax=0 be both 2/a and a/2?

Answer 65

plug them in and see

Answer 66

show me what you get if you are still confused

Answer 67

you should help highschool students here with their algebra and use wolfram alpha to prove stuff to yourself. Wolframalpha.com can do calculus for you to btw

Answer 68

2/a and a/2?

2-a(2/a) = 0
\[ 2-a^1 2a^{-1} = 0\]
add exponents of a
1 + (-1) = 1 - 1 = 0
\[ 2-a^0 2 = 0\]
\[ 2-(1)2 = 0\]

remember anything with the exponent 0 equals 1, excluding 0^0 which is undefined


as for the other thing you suggested

\[ 2-2\frac{a}{2} = 0\]
\[ 2- a^1a^1 = 0\]
\[ 2- a^2 = 0\]

we dont know if a = (2)^(1/2) is so yeah it isn't a definite critical point

Answer 69

well anyways I'm leaving hope I was helpful if this was implicit differentiation
check that link out, if it is calc 1 you will have to learn it anyways so might be worth checking out, you might also want to check khan academy or patrickmjt or what ever his name is on youtube

Answer 70

thank you sooooo much