Posted by **Aaron** on Saturday, December 1, 2012 at 9:49am.

d/dx[lnsquare root of(x^3) cube root of (x+3) fifth root of (3x-2)]

- Calculus -
**Reiny**, Saturday, December 1, 2012 at 10:48am
are we taking ln of the whole thing??

I read it as

d( ln [ (x^3)^(1/2) (x+3)^(1/3) (3x-2)^(1/5) ] /dx

let y = ln [ (x^3)^(1/2) (x+3)^(1/3) (3x-2)^(1/5) ]

remember ln(AB) = lnA + lnB

so our

y = ln (x^3)^(1/2) + ln (x+3)^(1/3) + ln (3x-2)^(1/5)

and remember ln A^n = n lnA

= 3(1/2) lnx + (1/3)ln(x+3) + (1/5)ln(3x-2)

dy/dx = (3/2)(1/x) + (1/3)(1/(x+3) + 1/5)(3/(3x-2))

= **3/(2x) + 1/(3(x+3)) + 3/(5(3x-2))**

I would not take it any further, it would look even messier, having to form a common denominator of

30(x+3)(3x-2)

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