Let us see (visually) how the derivative of $\sin_p$ with respect to $p$ looks like. Recall that the generalized $p$-trigonometric function $\sin_p(x)$ for $x \in (0,\pi_p/2)$ can be defined as the inverse of the function

\[\frac{\pi_p}{2} \frac{B(y^p,1/p,1-1/p)}{B(1/p,1-1/p)}, \quad y \in (0,1),\]

where $B(y^p,1/p,1-1/p)$ and $B(1/p,1-1/p)$ stand for the incomplete and complete Beta functions, respectively. See, e.g., Eq. (2.15) in [¹]. Here, $\pi_p = \frac{2\pi}{p \sin(\pi/p)}$.

Consider the function $\sin_p(\pi_p x/2)$ for $x \in (0,1)$ which is then defined as the inverse of

\[\frac{B(y^p,1/p,1-1/p)}{B(1/p,1-1/p)}, \quad y \in (0,1).\]

Using the relation between the derivatives (wrt a parameter) of a function and its inverse, and by launching, say, Mathematica, we can obtain the following figures. (Of course, if I didn’t mess up with the code). The blue graphs are the graphs of $\sin_p(\pi_p x/2)$ for $x \in (0,1)$, and the orange ones are the graphs of the corresponding derivative wrt $p$.

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Last modified: 23-Mar-22

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