Question

find the derivative of y=x(e^2)-e

Answer 1

It's

\[\Large y = x(e^2)-e\]

right?

Answer 2

yes!

Answer 3

e is just a number (it's roughly 2.71828...) it's like pi

Answer 4

so e^2 is just a number

Answer 5

so e^2 is effectively the coefficient of the x term

Answer 6

so if I asked you to derive y = 2x+1 for me, what would you say?

Answer 7

2

Answer 8

good, if you have a line, the derivative is just the slope of that line

Answer 9

I know the answer to the problem, I just don't see how to get to it

Answer 10

this is no different (it's just a bit more convoluted)

Answer 11

the slope in this case e^2, so the derivative of y is e^2

Answer 12

they say it's dy/dx = e^2 - e^x

Answer 13

oh then there has to be something more to the problem

Answer 14

so the problem can't be

\[\Large y = x(e^2)-e\]

Answer 15

If the problem was

\[\Large y = x(e^2)-e^x\]

then that would the correct answer

Answer 16

oh, it's that, yes, but how does that work :(

Answer 17

i'm so confused

Answer 18

you see how the derivative of x(e^2) is e^2 right?

Answer 19

yes!

Answer 20

so the only thing left is to take the derivative of e^x

this will give you e^x

Answer 21

this is why e is so special

Answer 22

OH, ok thank you! can I chat you if I have more questions?

Answer 23

it's because it is its own derivative

Answer 24

sure

Answer 25

and I'm sure you have learned this rule

if y = b^x

then

dy/dx = b^x*ln(b)


if b = e, then you get

y = e^x

dy/dx = e^x

Answer 26

because my whole work for the night is going to be derivatives of logs, ln, and e

Answer 27

we have all the rules, lnx = 1/x, lnu = 1/u du/dx, e^x = e^x, logau = 1/xlna, etc

Answer 28

yes, if

y = ln(x)

then

dy/dx = 1/x

etc etc

Answer 29

I have a question

Answer 30

there's a difference between logau and logax

Answer 31

a being the base

Answer 32

when do you use each formula, you use logax when there's an x as the variable right?

Answer 33

what were the formulas you are referring to again?

Answer 34

if y = log(a,x)

then

dy/dx = 1/(x*ln(a))

right?

Answer 35

log(a,x) means "log base 'a' of x"

Answer 36

i mean log base 'a' of x

Answer 37

but when do you use log base a of x and log base a of u?

Answer 38

i know

Answer 39

unless i can see what formulas you're working with, they both sound the same (just replace x with u)...so I see no difference

Answer 40

i think you mean

log(a,u) = 1/u*ln(a) * dy/dx

Answer 41

or to be more accurate

the derivative of log(a,u) with respect to x is 1/( u*ln(a) ) * du/dx

Answer 42

then what did he mean with log (x, u) equaling that without du/dx?

Answer 43

log(x, u) means log base x of u

shouldn't it be log(a, u) or log(a, x)

Answer 44

i meant, log (a, x)

Answer 45

seriously having a mental crash right now, it's all jumbling, nothings making sense, and the tests tomorrow

Answer 46

have you heard of the chain rule?

Answer 47

yeah!

Answer 48

that's what's going on

normally, the derivative of log(a,x) is just \[\Large \frac{1}{x\ln(a)}\]

Answer 49

but when you introduce the 'u' in there, it complicates things because you have to use the chain rule

Answer 50

does that make sense?

Answer 51

what's the difference between a, and u?

Answer 52

*x and u

Answer 53

when it says x does it mean, literally just the variable x

Answer 54

or could you put log 4 x^2 and stilll use log a x

Answer 55

or would the x^2 be used as u

Answer 56

u and x are their own variables, but when we say dy/dx we mean "derive the function y with respect to x"

So...

if you have something with u in it (and no x), then either u is assumed to be a constant (not likely though) or it's a function of x on its own

ex:

if u = 2x

and y = log(3, u)

this is the same thing as saying

y = log(3, 2x)

Answer 57

this is why it's a bit more complicated

Answer 58

so log 5 x you would use log a x

Answer 59

or something with a constant base, and x

Answer 60

but you use log a u for anything pretty much?

Answer 61

brb going to shower, to clear the headache

Answer 62

yes, log(a, x) allows you to say log(5, x), but you can't stray from that (so only 'a' is allowed to change)

but

log(a, u) can be more difference since you can replace 'u' with any expression of x

Answer 63

different*