anudoneddbv

2022-07-16

Considering s is implicit to function of $p$, given by ${s}^{6}-{p}^{4}=1$. For what s is it increasing and decreasing?
Well, I answeblack first like following: Calculating the first derivative using the implicit differentiation is ${s}^{\prime }=\frac{2{p}^{3}}{3{s}^{5}}$. $s$ is increasing where ${s}^{\prime }>0$ and decreasing where ${s}^{\prime }<0$.
Obviously, I answeblack too generally and I'm wondering how I better answer it using the first derivative to support my answer.

Expert

We are talking here about the solution set

where

The implicit function theorem says the following: When $\left({p}_{0},{s}_{0}\right)\in S$ and the "technical condition"
$\begin{array}{}\text{(1)}& \frac{\mathrm{\partial }F}{\mathrm{\partial }s}{|}_{\left({p}_{0},{s}_{0}\right)}=6{s}_{0}^{5}\ne 0\end{array}$
is satisfied then there is a window

and a ${C}^{1}$-function

such that $S\cap W$ equals the graph of $\psi$. Furthermore one has

Now $S$ contains no points with $s=0$; therefore (1) is satisfied at all points of $S$S.

Whether the local function $\psi$ is increasing or decreasing in its window depends on the signs of ${p}_{0}$ and ${s}_{0}$: When ${p}_{0}$ and ${s}_{0}$ have the same sign $\psi$ is increasing (for small enough $h>0$), otherwise $\psi$ is decreasing. The points $\left(0,±1\right)\in S$ are special, since ${\psi }^{\prime }\left(0\right)=0$ there. Here further analysis is necessary, which I leave to you.

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