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April 18, 2014

April 18, 2014

Posted by **Marissa** on Tuesday, August 26, 2008 at 11:44pm.

Differentiate the following:

1. y = ln [(4-x)/(4+x)]

Since it's the ln of the entire thing, then I can apply ln a - ln b

so, dy/dx = y' = 1/(4-x) *(-1) - 1/(4+x) *(1)

dy/dx = y' = -1/(4-x) - 1/(4+x)

2. y = ln √((3u+2)/(3u-2))

so that's the same as

y = ln ((3u+2)/(3u-2))^(1/2)

= 1/2 ln ((3u+2/3u-2))

using the ln a - ln b property again:

1/2[ln(3u+2) - ln(3u-2)]

dy/dx = 1/2 [1/(3u+2)*3 - 1/(3u-2)*3]

dy/dx = 1/2[3/(3u+2) - 3/(3u-2)

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/(3t-1)^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)][(3t-1)^-4]

It would still be messy to simplify.

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