Math differentiation

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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 [(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 -

    #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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