# Formula:KLS:14.10:21

${\displaystyle{\displaystyle{\displaystyle D_{q}\left[{\tilde{w}}(x;q^{\alpha}% ,q^{\beta}|q)\ctsqJacobi{\alpha}{\beta}{n}@{x}{q}\right]{}=-2q^{-\frac{1}{2}% \alpha+\frac{1}{4}}\frac{(1-q^{n+1})(1+q^{\frac{1}{2}(\alpha+\beta-1)})(1+q^{% \frac{1}{2}(\alpha+\beta)})}{1-q}{}{\tilde{w}}(x;q^{\alpha-1},q^{\beta-1}|q)P_% {n+1}^{(\alpha-1,\beta-1)}(x|q)}}}$

## Substitution(s)

${\displaystyle{\displaystyle{\displaystyle{\tilde{w}}(x;q^{\alpha},q^{\beta}|q% ):=\frac{w(x;q^{\alpha},q^{\beta}|q)}{\sqrt{1-x^{2}}}}}}$ &

$\displaystyle {\displaystyle w(x):=w(x;q^{\alpha},q^{\beta}|q) =\left|\frac{\qPochhammer{\expe^{2\iunit\theta}}{q}{\infty}} {\qPochhammer{q^{\frac{1}{2}\alpha+\frac{1}{4}}\expe^{\iunit\theta},q^{\frac{1}{2}\alpha+\frac{3}{4}}\expe^{\iunit\theta} -q^{\frac{1}{2}\beta+\frac{1}{4}}\expe^{\iunit\theta},-q^{\frac{1}{2}\beta+\frac{3}{4}}\expe^{\iunit\theta}}{q}{\infty}}\right|^2 =\left|\frac{\qPochhammer{\expe^{\iunit\theta},-\expe^{\iunit\theta}}{q^{\frac{1}{2}}}{\infty}}{\qPochhammer{q^{\frac{1}{2}\alpha+\frac{1}{4}}\expe^{\iunit\theta} -q^{\frac{1}{2}\beta+\frac{1}{4}}\expe^{\iunit\theta}}{q^{\frac{1}{2}}}{\infty}}\right|^2 =\frac{h(x,1)h(x,-1)h(x,q^{\frac{1}{2}})h(x,-q^{\frac{1}{2}})} {h(x,q^{\frac{1}{2}\alpha+\frac{1}{4}})h(x,q^{\frac{1}{2}\alpha+\frac{3}{4}}) h(x,-q^{\frac{1}{2}\beta+\frac{1}{4}})h(x,-q^{\frac{1}{2}\beta+\frac{3}{4}})}}$ &
${\displaystyle{\displaystyle{\displaystyle h(x,\alpha):=\prod_{k=0}^{\infty}% \left(1-2\alpha xq^{k}+\alpha^{2}q^{2k}\right)=\qPochhammer{\alpha e^{i\theta}% ,\alpha e^{-i\theta}}{q}{\infty}}}}$ &

${\displaystyle{\displaystyle{\displaystyle x=\cos\theta}}}$

## Proof

We ask users to provide proof(s), reference(s) to proof(s), or further clarification on the proof(s) in this space.