Question

please help ive got an exam tomorrow: determine the equation of the line perpendicular to f(x) = 0.5e^(x+1) at its y-int

Answer 1

at which point on f(x)?

Answer 2

at its y-int

Answer 3

the y-int is when x=0

Answer 4

plug in x = 0 into f(x) and you get which output value?

Answer 5

i get 1.36

Answer 6

you sure? try it again

Answer 7

oh wait, had a diff value programmed into my calculator for e (weird)

Answer 8

hold on

Answer 9

yes i tried it again i still get 1.36

Answer 10

ok yes I'm now getting that

Answer 11

so the y-intercept is roughly (0,1.36)

Answer 12

what is the derivative of f(x) ?

Answer 13

i get \[f'(x) = (x+1) e ^{x+1}(\frac{ 1 }{ 2 }) \]

Answer 14

but i'm not sure if its right

Answer 15

you're using the power rule when you should be using logarithmic differentiation

Answer 16

have you heard of that method before?

Answer 17

no...but we have been differentiating log. formulas...i'm guessing that is the method?

Answer 18

ok one sec

Answer 19

ok hopefully this page looks familiar to you then (the topics and ideas look familiar)

https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/logdiffdirectory/

Answer 20

so the deriviative would be (x+1)e^x?

Answer 21

no

Answer 22

f(x) = 0.5e^(x+1)

y = 0.5e^(x+1)

ln( y ) = ln( 0.5e^(x+1) )

ln( y ) = (x+1)ln( 0.5e )

ln( y ) = (x+1)( ln( 0.5) + ln( e ) )

ln( y ) = (x+1)( ln( 0.5) + 1 )

d/dx[ ln( y ) ] = d/dx[ (x+1)( ln( 0.5) + 1 ) ]

y'/y = ln( 0.5) + 1

y' = y*(ln( 0.5) + 1)

y' = 0.5e^(x+1)*(ln( 0.5) + 1)

or something along those lines

Answer 23

Huh? since when is \[\frac{d}{dx}[e^{x-1}]\ne e^{x-1}\]?

Answer 24

Sorry, you've got \[\frac{d}{dx}[e^{x+1}]\]but the question stands...

Answer 25

hmm I guess you could do this

f(x) = 0.5e^(x+1)

f(x) = 0.5e^x*e^1

f(x) = 0.5e*e^x

f ' (x) = 0.5e*e^x

Answer 26

that's probably a lot easier

Answer 27

\[f'(x) = f(x) = \frac{1}{2}e^{x+1}\]Agreed?

Evaluate that at x = 0 to get the instantaneous slope. take the negative reciprocal to find the slope of a perpendicular line, and fit a line through the y-intercept. Am I missing anything?

Answer 28

no that works too, I just was thinking of something else

Answer 29

f ' (x) = 0.5e*e^x ---> f ' (x) = 0.5*e^(x+1)

Answer 30

how is the derivative equal to the same thing as the function?

Answer 31

but logarithmic differentiation still works here (at least it should)

Answer 32

the magic of e :-)

Answer 33

derivative of e^x is e^x

Answer 34

it does work of course, if you don't make any errors in the process...

Answer 35

yes but isn't the derivative of e^bx = (b)e^bx

Answer 36

or something like that

Answer 37

you're thinking of the chain rule

Answer 38

f(x) = e^(g(x))

f ' (x) = g ' (x) *e^(g(x))

Answer 39

if g(x) = x, then g ' (x) = 1

Answer 40

so the chain rule isn't applied when e^x+1 correct?

Answer 41

it can be, but it's not really necessary

Answer 42

if you are having a test, you should learn
that
\[ \frac{d}{dx} e^{f(x)} = e^{f(x)} \frac{d}{dx} f(x) \]

Answer 43

f(x) = e^(x+1)

f ' (x) = e^(x+1) * d/dx[x+1]

f ' (x) = e^(x+1) * 1

f ' (x) = 1*e^(x+1)

f ' (x) = e^(x+1)

Answer 44

in this case you will get back the original e^(x+1) because d/dx (x+1) = 1

Answer 45

ohhhhhh

Answer 46

so what about the 0.5...shouldn't the product rule be used?

Answer 47

0.5 is a constant you can pull out

Answer 48

you can put any constant multipliers "out front"

Answer 49

like always: d/dx 2x is 2 * d/dx x

Answer 50

so any number infront of e would be considered a constant?

Answer 51

if it is a constant. if it is a variable, you would use the product rule

Answer 52

ohhh i see. so if there is a cos infron of the e it would be pulled out but if there was a cosx infront of the e than it would be variable and we would use the product rule correct?

Answer 53

yes if by cos you mean for example cos(pi) where the angle is constant

Answer 54

yes

Answer 55

lol sorry for all the dumb question but i really need to make this clear for the exam

Answer 56

I assume you can finish this problem?

Answer 57

yes u can thank you both for ur help!

Answer 58

*i loool awks typo