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 ′ = 2 p 3 3 s 5 . s is increasing where s ′ &gt; 0 and decreasing where s ′ &lt; 0.Obviously, I answeblack too generally and I'm wondering how I better answer it using the first derivative to support my answer.

Question

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    ′    =                    2                  p          3                            3                  s          5                    .   s is increasing where       s    ′    &amp;gt;  0 and decreasing where       s    ′    &amp;lt;  0.Obviously, I answeblack too generally and I&#039;m wondering how I better answer it using the first derivative to support my answer.

slapadabassyc · Accepted Answer

We are talking here about the solution set  S  :=            {        (  p  ,  s  )              |          F  (  s  ,  p  )  =  0            }           ,where  F  (  p  ,  s  )  =      s    6    −      p    4    −  1     .The implicit function theorem says the following: When   (      p    0    ,      s    0    )  ∈  S and the &quot;technical condition&quot;                    (1)                                                        ∂              F                                      ∂              s                                                                          |                                            (                          p              0                        ,                          s              0                        )                          =        6                  s          0          5                ≠        0            is satisfied then there is a window  W  :=     ]      p    0    −  h  ,      p    0    +  h  [     ×     ]      s    0    −      h    ′    ,      s    0    +      h    ′    [   and a       C    1  -function  ψ  :    ]      p    0    −  h  ,      p    0    +  h  [     →     ]      s    0    −      h    ′    ,      s    0    +      h    ′    [     ,such that   S  ∩  W equals the graph of   ψ. Furthermore one has      ψ    ′    (      p    0    )  =  −                              F          p                (                  p          0                ,                  s          0                )                              F          s                (                  p          0                ,                  s          0                )              =                    2                  p          0          3                            3                  s          0          5                         .Now   S contains no points with   s  =  0; therefore (1) is satisfied at all points of   SS.Whether the local function   ψ 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   ψ is increasing (for small enough   h  &amp;gt;  0), otherwise   ψ is decreasing. The points   (  0  ,  ±  1  )  ∈  S are special, since       ψ    ′    (  0  )  =  0 there. Here further analysis is necessary, which I leave to you.