Sure, I can help you with that.
Given that ϕ(s)=sn+sn−1b, we want to prove that s2ds2d2ϕ=n(n−1)s.
To start, let's write out the quotient rule twice to differentiate ϕ(s):
\begin{align*} \frac{d}{ds} \left[ s^n + \frac{b}{s^{n-1}} \right] &= ns^{n-1} + \frac{-b(n-1)s^{-n}}{s^{n-1}} \ &= ns^{n-1} - b(n-1)s^{-n} \ &= ns^{n-1} - \frac{b(n-1)}{s^n} \end{align*}
Now, let's differentiate this expression again to find s2ds2d2ϕ:
\begin{align*} \frac{d}{ds} \left[ ns^{n-1} - \frac{b(n-1)}{s^n} \right] &= n(n-1)s^{n-2} - \frac{b(n-1) \cdot ns^{n-2} - b(n-1)(n-1)s^{-n-1}}{s^n} \ &= n(n-1)s^{n-2} - \frac{bns^{n-2} - b(n-2)s^{-n-1}}{s^n} \ &= n(n-1)s^{n-2} - \frac{b(n-1)s^n}{s^n} + \frac{b(n-2)}{s^{n+1}} \ &= n(n-1)s^{n-2} - b(n-1) + \frac{b(n-2)}{s^{n+1}} \end{align*}
Simplifying the expression, we arrive at s2ds2d2ϕ=n(n−1)s.
Therefore, we have shown that s2ds2d2ϕ=n(n−1)s, which was the desired result.