Posted by Marissa on Tuesday, August 26, 2008 at 11:44pm.
Can somebody please check these over for me? I really want to know what I did wrong if I made mistakes!
Differentiate the following:
1. y = ln [(4x)/(4+x)]
Since it's the ln of the entire thing, then I can apply ln a  ln b
so, dy/dx = y' = 1/(4x) *(1)  1/(4+x) *(1)
dy/dx = y' = 1/(4x)  1/(4+x)
2. y = ln √((3u+2)/(3u2))
so that's the same as
y = ln ((3u+2)/(3u2))^(1/2)
= 1/2 ln ((3u+2/3u2))
using the ln a  ln b property again:
1/2[ln(3u+2)  ln(3u2)]
dy/dx = 1/2 [1/(3u+2)*3  1/(3u2)*3]
dy/dx = 1/2[3/(3u+2)  3/(3u2)
3. y = f(x) = tan(ln(4x+1))
so first I found the derivative of ln 4x+1
= 1/(4x+1)*4 = 4/(4x+1)
I plugged that number back into the original, making it
y = f(x) = tan(4/(4x+1))
Since the d/dx of tan x = sec^2x,
does it just end up being
dy/dx = sec^2(4/(4x+1)) ?
I am confused on how to do a problem like this:
4. f(t) = ln (2t+1)^3/(3t1)^4
I can't apply the ln a  ln b property since it's not the ln of the entire thing, so how should I do it?

Math differentiation  Reiny, Wednesday, August 27, 2008 at 12:57am
#1 and #2 are correct
in #3
if y = tan(u)
then y' = sec^2 (u) * du/dx
so y = tan(ln(4x+1))
y' = sec^2 [ln(4x+1)]*4/(4x+1)
#4 is a messy quotient rule problem
perhaps you could use the product rule by rewriting it as
f(t) = [3ln (2t+1)][(3t1)^4]
It would still be messy to simplify.