Question

how can i do the derivates of
(x+1)^3/x3/2

Answer 1

Is this the function?

\[(x + 1)^3/x ^{3/2}\]

Answer 2

yep

Answer 3

Ok... rewrite:

\[y = (x + 1)^3x ^{-3/2}\]
Product rule:
\[y'= 3(x + 1)^2(1)x ^{-3/2}+(x+1)^3(-3/2)x ^{-5/2}\]
\[y'=(x+1)^2x ^{-3/2}[3+(-3/2)(x+1)x ^{-1}]\]
\[y'=(x+1)^2[3/2-3/(2x)]/x ^{3/2}\]
\[y'=(x+1)^2[(3x-3)/(2x)]/x ^{3/2}\]
\[y'=3(x-1)(x+1)^2/(2x ^{5/2})\]

Someone else may wanna check this over, though...

Answer 4

Yup I checked. It's definitely right.

Answer 5

i got lost on the second line

Answer 6

On the second line I factored out (x + 1)^2 and x^(-3/2).