Can someone please explain this? (Log and Exp) (1 Viewer)

[LuthienAdrianna]

Here's an example from my textbook:

Find the derivative of 2^x:

2 = eln2

Therefore, 2^x = (eln2)^x = exln2

dy/dx = ln2exln2 = ln2 . 2^x = (2^x) ln 2

I don't get it. ><

 

[Dreamerish*~]

Let y = 2^x

log_ey = log_e2^x

= xlog_e2

x = (log_ey)/(log_e2)

= 1/log_e2 * log_ey

dx/dy = 1/log_e2 * 1/y

= 1/ylog_e2

Flip it:

dy/dx = ylog_e2

Since y = 2^x

dy/dx = 2^xlog_e2

Look at this thread: http://community.boredofstudies.org/showthread.php?t=89421. (Thanks to those who explained it )

 

[Antwan23q]

lol
just use

y=2^x
lny = ln(2^x)
lny = xln2
lny/ln2 = x (do the derivative with respect to y, dx/dy)

1/yln2 = dx/dy (then u flip it)

dy/dx = yln2

and y = 2^x
so dy/dx = (2^x)ln2

 

[LuthienAdrianna]

Thanks! XD

ETA: Why did you use dx/dy? I've not come across that before :\

 

[icycloud]

Thanks! XD

ETA: Why did you use dx/dy? I've not come across that before :\

That simply means the derivative of x with respect to y. Note that dy/dx = 1/(dx/dy)

 

[KFunk]

I reckon the 'e' method makes life easier. Luthian, in your first post it's just subbing in e^ln2 for 2 because 2^x = e^x.ln2 so your working is:

y = 2^x = (e^ln2)^x = e^x.ln2

dy/dx = ln2.e^x.ln2 = ln2.2^x *

(* because e^x.ln2 = 2^x as was shown)

 

[LuthienAdrianna]

*has a moment of clarity*

Thanks! LOL maybe I should wait until we cover things in class before attempting them myself.

 

[KFunk]

In case it helps this is why all that stuff works:

log_e2 = log_e2 (which is obvious)

log_e2 = (log_e2)(log_ee) (because log_ee = 1)

log_e2 = log_ee^ln2 (using log laws)

&there4; 2 = e^ln2 (removing the logs)


So that's the logic behind that substitution. It's a good trick to use because when you have something in the form e^kx you know right away that the derivative is ke^kx.