Question

problem with a second derivative.
I will ask the question in the next post

Answer 1

problem - find second derivative of
\[y = (1 + \frac{1}{x})^{3}\]
It seems simple, but it quickly turned complicated, so I replaced
\[u = (1 + \frac{1}{x})\]\[u' = -x^{-2}\]\[u'' = 2x^{-3}\]

Then (and please check if this is correct):
\[y' = 3u^2u'\]\[y'' = 3u^2u'' +u' 6u u''\]
And when I replace 'u' and simplify I end up with
\[(\frac{6}{x^3})(1 + \frac{3}{x} + \frac{2}{x^2})\] which I can change to
\[(\frac{6}{x^4})(1 + \frac{1}{x})(1+\frac{2}{x}),\]
but the answer in the book is
\[(\frac{6}{x^3})....\]
so I have one extra 'x' in the denominator.
So the question: Is my expression for y'' correct and I just made a mistake somewhere (I have checked it several times), or am I missing something in the expression for y' and/or y''?

Answer 2

oops - I mean\[y'' = 3u^2u'' + u'6uu'\]

Answer 3

Your way of doing it seems to be more complicated. Simplest is to write it as y = 1 + 3x^-1 + 3x^-2 + x^-3 and go from there.

Answer 4

I found my mistake, and I do get \[\frac{6}{x^3}(...)\]