edited, forget it.

 

Did you get the help you needed, or simply decide that there are more important things in life than calculus? (Like number theory, algebraic topology, . . . .)

 

[QUOTE]Originally posted by magician
[B]Did you get the help you needed, or simply decide that there are more imprtant things in life than calculus?

 

post it

 

I believe he means "post it, please"

 

hehe, ya

 

aw man ok fine, but it makes no difference anymore

btw, why can't we delete our own threads anymore? *shrug*

f(x) = 2x + lnx

find g'(2)

FYI:
inverse function

g'(x) = 1/f'(g(x))

note to anyone that may have forgotten

g(x) is NOT the reciprocol of f(x), g(x) is obtained when you switch the x and the y and solve for y...

i tried that method and couldn't get it, i can't get g(x).

 

hmm, have you tried using the anti-derivative maybe?

 

If I understand you, g(x) is the inverse of f(x): f(g(x)) = g(f(x)) = x.

In that case, if y = f(x), so x = g(y), then g'(y) = 1 / f'(x). If you think of slopes and exchange the roles of x and y this reciprocal relationship is pretty clear: if y changes by 2 when x changes by 1 then dy/dx = 2, and dx/dy = 1/2.

Therefore, if x = g(2) then 2 = f(x); solve this for x, getting x = 1.

f'(x) = 2 + 1 / x, so f'(1) = 2 + 1 / 1 = 3.

g'(2) = 1 / f'(1) = 1/3.

Voila!

(Note that this can be done without an explicit formula for g(x).)

 

[QUOTE]Originally posted by magician
[B]If I understand you, g(x) is the inverse of f(x): f(g(x)) = g(f(x)) = x.

In that case, if y = f(x), so x = g(y), then g'(y) = 1 / f'(x).