Let $k \in (0,1)$ and $w>0$. Consider the function $\varphi: \mathbb{R} \longrightarrow \mathbb{R}$ given by $$\varphi(\xi)= \frac{\sqrt{2}k}{\sqrt{k^2+1}}\cdot \text{sn} \left(\frac{\xi}{\sqrt{w}\sqrt{k^2+1}};k\right),\forall \ \xi \in \mathbb{R},$$ where $\text{sn}$ denote the Jacobi elliptic function snoidal. I want calculate $\eta:=\frac{d}{dw} \varphi$. How do I proceed? Do I apply the chain rule normally? In doing so, I arrived at $$\eta (\xi)= \frac{\sqrt{2} \;k}{\sqrt{k^2+1}} \frac{\xi}{\sqrt{k^2+1}}\cdot\left(\frac{-1}{2\sqrt{w^3}}\right)\cdot \frac{d}{d\xi} \text{sn}\left(\frac{\xi}{\sqrt{w}\sqrt{k^2+1}};k\right), \; \forall \; \xi \in \mathbb{R}.$$
Is that correct? If not, how should I proceed?
First: $$ \frac{d}{dz} \operatorname{sn}(z;k) = \operatorname{cn}(z;k)\operatorname{dn}(z;k). $$ Then $$ \frac{d}{dw}\left[{\frac {\sqrt {2}k}{\sqrt {{k}^{2}+1}}{\;\operatorname{sn}} \left( {\frac { \xi}{\sqrt {w}\sqrt {{k}^{2}+1}}};k \right) }\right] =\\ -\,{\frac {\sqrt {2}k\xi}{2 \left( {k}^{2}+1 \right) {w}^{3/2}}{\;\operatorname{cn}} \left( {\frac {\xi}{\sqrt {w}\sqrt {{k}^{2}+1}}};k \right) { \;\operatorname{dn}} \left( {\frac {\xi}{\sqrt {w}\sqrt {{k}^{2}+1}}};k \right) } $$
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When you write $\frac{d}{d\xi}$ in there, it is confusing. But in one interpretation, your calculation could be correct.