Question

Find the derivative of

f(x) = 1/2 (e^x + e^-x)

Answer 1

haha, i just did this in class today, just a second

Answer 2

do you want how to get it or just the answer?

Answer 3

If you want how to get it I will toss it into the software I have so you know step by step, just let me know. Heres the answer though:

f'(x)=(e^(-x)(e^(2x)-1))/(2)

Answer 4

Could you show me how you got it.

Answer 5

sure, I just looked over it and we got it the same way, where is this problem from by the way?

Answer 6

f(x)=(1)/(2)*(e^(x)+e^(-x))

The derivative of f(x) is equal to f'(x)=(d)/(dx) (1)/(2)*(e^(x)+e^(-x)).
f'(x)=(d)/(dx) (1)/(2)*(e^(x)+e^(-x))

Multiply the rational expressions to get ((e^(x)+e^(-x)))/(2).
(e^(x)+e^(-x))/(2)

Find the derivative of the expression.
(d)/(dx) (e^(x)+e^(-x))/(2)

The chain rule states that the derivative of a composite function (f o g)' is equal to (f' o g)*g'. To find the derivative of ((e^(x)+e^(-x)))/(2), find the derivatives of each portion of the function and use the chain rule formula.
(d)/(du) (u)/(2)*(d)/(dx) e^(x)+e^(-x)

To find the derivative of (u)/(2), multiply the base (u) by the exponent (1), then subtract 1 from the exponent (1-1=0). Since the exponent is now 0, u is eliminated from the term.
(d)/(du) (u)/(2)=(1)/(2)

The derivative of e^(x) is e^(x). This is true because (d)/(dx) a^(x)=a^(x)ln(a). In this case, the base is e and ln(e)=1, so the derivative is e^(x).
(d)/(dx) e^(x)+e^(-x)=e^(x)+(d)/(dx) e^(-x)

To find the derivative of -x, multiply the base (x) by the exponent (1), then subtract 1 from the exponent (1-1=0). Since the exponent is now 0, x is eliminated from the term.
(d)/(dx) -x=-1

Using the chain rule, the derivative of e^(-x) is e^(-x)*-1.
e^(-x)*-1

Multiply e^(-x) by -1 to get -e^(-x).
-e^(-x)

The derivative of e^(x)+e^(-x) is e^(x)-e^(-x).
(d)/(dx) e^(x)+e^(-x)=e^(x)-e^(-x)

Replace the variable u with e^(x)+e^(-x) in the expression.
(d)/(du) (u)/(2)=(1)/(2)

Replace the variable u with e^(x)+e^(-x) in the expression.
e^(x)-e^(-x)

Form the derivative by substituting the values for each portion into the chain rule formula.
=(1)/(2)*(e^(x)-e^(-x))

Multiply the rational expressions to get ((e^(x)-e^(-x)))/(2).
(d)/(dx) (e^(x)+e^(-x))/(2)=(e^(x)-e^(-x))/(2)

Factor out the GCF of e^(-1x) from the expression e^(x).
(d)/(dx) (e^(x)+e^(-x))/(2)=(e^(-1x)(e^(2x))-e^(-x))/(2)

Factor out the GCF of e^(-1x) from the expression -e^(-x).
(d)/(dx) (e^(x)+e^(-x))/(2)=(e^(-1x)(e^(2x))+e^(-1x)(-1))/(2)

Factor out the GCF of e^(-1x) from e^(x)-e^(-x).
(d)/(dx) (e^(x)+e^(-x))/(2)=(e^(-1x)(e^(2x)-1))/(2)

Remove the -1 from -1x.
(d)/(dx) (e^(x)+e^(-x))/(2)=(e^(-x)(e^(2x)-1))/(2)

The derivative of the function is (e^(-x)(e^(2x)-1))/(2).
f'(x)=(e^(-x)(e^(2x)-1))/(2)

Answer 7

?

Answer 8

he asked how to get it, Ihave a software program that generate step by step...with detailed explanations. Makes it easier than typing it all..lol

Answer 9

why not
\[\frac{1}{2}(e^x-e^{-x})\]

Answer 10

if that is what he means satellitethen it is just the integrall of sinh(x)

Answer 11

see. I thought about that, but my instructore earlier today worked the same exact problem, and this is the direction he went too. not sure whats up.

Answer 12

am i lost? maybe tired. but there is nothing easier than finding the derivative of
\[e^x\] it is just
\[e^x\] and likewise the derivative of
\[e^{-x}\] is
\[-e^{-x}\] so you are done right away

Answer 13

this does ask for "derivative" right? not "integral"

Answer 14

yes but e^x-e^-x/2 is an identity for sinh(x)

Answer 15

even this: f(x) = 1/2 (e^x + e^-x), see I agree with you. satellite

Answer 16

yes sir, derivative

Answer 17

so you could do dy/dx if the 1/2 is out in front

Answer 18

that is what i did, and then applied the chain rule

Answer 19

yes but if you knew that the identity e^x-e^-x /2 = sinh(x).. you could just do the dy/dx(sinh(x)) = cosh(x)

Answer 20

yes if you wanted to prove that it was true, this would be the proof

Answer 21

stick it to the teacher... make her thinkyou know more than what you should lmfao

Answer 22

s quick quesiton, is the way I did it right or wrong?

Answer 23

it seems like one step to me i suppose you use the "chain rule" to find the derivative of
\[e^{-x}\]

Answer 24

i don't know it if is right or wrong but it is awfully long to read. the derivative of e^x is e^x
the derivative of e^(-x) is -e^(-x) and the one half is a red herring

Answer 25

hey sat, if you don't mind, tell me exactly how come its one step? by the way, we haven't even made it to the chain rule in derrivvatives this semester, I just used it anyway.

Answer 26

nm, u just told mehaha

Answer 27

is this waht you have?
\[\frac {dy}{dx}[\frac{1}{2(e^x-e^-x)}\]

Answer 28

if so you would have to use chain rule because you pull 1/2 out and then you are left with (e^x-e^-x)^-1

Answer 29

yes, thats it I think, I done put the notes away. i just posted all that to help him

Answer 30

we had another one like it, hold on ill pull it up, it was intense, but I don't think it was necessary to do everything we ddid in that one either.

Answer 31

or was it\[\frac{dy}{dx}[\frac{1}{2}e^x-e^-x]\]

Answer 32

you did it right lawnphysics since (e^(-x)(e^(2x)-1))/(2)=(e^x-e^(-x))/2 is an identity (verified by wolfram alpha)


But outkast3r09 has a good point


\[\Large \frac{1}{2}(e^x+e^{-x})=\cosh(x)\]


and \[\Large \frac{d}{dx}(\cosh(x)) = \sinh(x)=\frac{1}{2}(e^x-e^{-x})\]


So \[\Large \frac{d}{dx}\left(\frac{1}{2}(e^x+e^{-x})\right)=\frac{1}{2}(e^x-e^{-x})\]

Answer 33

so if you can see this, then it saves a lot of work

Answer 34

i see both, thank you, that just helps me more. came in here to answer a question andlearned something too lol

Answer 35

^ yes and lawnphysics, this stuff isn't really taught until integrals but it would help to get reps in right now too instead of taking this beyond what it needs to be

Answer 36

yeah, but my instructor is hard core he don't like us using stuff outside of what he has alreadt tuahgt

Answer 37

and the derivative isn't that bad also

pull out one half and simply use the additive property so you have 1/2(dy/dx[e^x)+dy/dx[e^-x)

dy/dx[e^u]=e^u u' so your just going to get e^x-e^-x

Answer 38

here ill show you how to get it

Answer 39

i hate how slow this thing types, has anyone ever though about a chat service just to speed things uP?

Answer 40

See what you think of this one outkast3r09

Answer 41

here ill show you how to get it
\[\frac{dy}{dx}[\frac{1}{2}(e^x-e^-x)=\frac{1}{2}[\frac{dy}{dx}e^x-\frac{dy}{dx}e^-x]\]

Answer 42

\[\frac{dy}{dx}[e^u]=e^uu'\]

Answer 43

then simply solve

Answer 44

now i know why he did it the way he did, the directions specifically say to use product or quotient rule then rewrite algebraically and apply quotient rule

Answer 45

but knowing easier ways makes me happy anyway haha

Answer 46

@jimthompson
i must be getting tired, but what is the work in finding the derivative of e^x?

Answer 47

dy/dx[e^u] = e^u u'

Answer 48

I was referring to the lengthy steps lawnphysics wrote out

Answer 49

so e^x is simple e^x

Answer 50

and e^-x will be -e^-x

Answer 51

i didn't write them out, Calculus solved did

Answer 52

oh i see

Answer 53

oh ok. still don't even see the need for sinh or cosh. it is two seconds to solve. the only tiny sophistication is knowing that the derivative of
\[e^{-x}\] is
\[-e^{-x}\]

Answer 54

because it's much easier to just know that it's sinh(x) because the dy/dx of it is simple... +cosh(x)

Answer 55

i use it to quickly find steps if I ever get stuck on a problem, not to cheat like most people. it helps if your working at 3am and there is no one to call haha

Answer 56

and in that case you know that it's going to equal simple 1/2(e^x+e^-x)

Answer 57

hey sat, do you use any instant messenging services?

Answer 58

@outcast, why is the derivative of sinh cosh? because the derivative you just found shows it

Answer 59

no not really. google chat sometimes

Answer 60

i just use trillian, i hate how slow this things types haha

Answer 61

the derivatives of hyperbolic functions are positive while their inverses are not i believe

Answer 62

btw for your question lawnphysics the easier way (to me) is to write your denominator as (t^2-1)^-1

Answer 63

and use the product rule

Answer 64

you mean chain rule?

Answer 65

well, the thing you have to rememver outkast, is we have only learned basic der. in class right now we are stuck using the basics. i understand there are better ways to do it too. can't want to be allowed to. i think ders. are freaking crazy fun

Answer 66

\[\frac{d}{dt}[\frac{xt-4}{t^2-x}]=\frac{d}{dt}[(xt-4)(t^2-x)^{-1}\]

Answer 67

jim, you would have to use the chain ffor the second part

Answer 68

i just personally despise the quotient rule

Answer 69

i must be looking at the wrong part

Answer 70

oh ok I see now

Answer 71

here, i despised it so much, i just took pictures of the board because I didn't want to write it down...here is what we did...i will post them...haha

Answer 72

does anyone want to see them?

Answer 73

i think i'm alright so did you need help on that ?

Answer 74

?

Answer 75

nope, i was just saying thats how I learned to do the other one the original poster solved. and thats why I went that way with it

Answer 76

just having a discusion about it with ya :)

Answer 77

Christ almighty. What a lot of heat for so little light.

@Sat73 is the one with the neon light here.

Answer 78

well i think that the question was different from the question we answered still

Answer 79

thnx. i thought i was losing my mind

Answer 80

people need to use the equation editor

Answer 81

because if the e's are in the bottom the problem becomes much longer like his teachers