$$2nJ_{n}(x) = xJ_{n+1}(x) + xJ_{n - 1}(x)$$ $$\because 2nJ_{n}(x) = xJ_{n+1}(x) + xJ_{n-1}(x)$$ $$\therefore J_{n+1}(x) = \frac{2n}{x}J_{n}(x) - J_{n-1}(x)$$

Put n = $\frac{1}{2}$

$$J_{\frac{3}{2}}(x) = \frac{1}{x}J_{\frac{1}{2}}(x) - J_{\frac{-1}{2}}(x)............(A)$$ We know the particular solution of Bessel's function is $$J_{n}(x) = \sum\limits_{m=0}^{\infty}\frac{(-1)\bigg(\frac{x}{2}\bigg)^{2m + n}}{m! \sqrt{n + n + 1}} = x^n \sum\limits_{m = 0}^{\infty}\frac{(-1)^m x^{2m}}{2^{2m + n} m! \sqrt{m + n + 1}}$$ Put, $$n = \frac{1}{2}...(inJ_{n}(x))$$ $$\therefore J_{\frac{1}{2}}(x) = \sqrt{x} \sum\limits_{m = 0}^{\infty}\frac{(-1)^m x^{2m}}{2^{2m + n} m! \sqrt{m + \frac{1}{2} + 1}}$$ $$J_{\frac{1}{2}}(x) = \sqrt{\frac{x}{2}} \sum\limits_{m = 0}^{\infty} \frac{(-1)^m x^{2m}}{2^{2m} m! \sqrt{m + \frac{3}{2}}}....(B)$$ Now, Is $$\sqrt{m + \frac{3}{2}} = \bigg(m + \frac{1}{2}\bigg)\bigg(m - \frac{1}{2}\bigg)\bigg(m - \frac{3}{2}\bigg)......\frac{3}{2} * \frac{1}{2} \sqrt{\frac{1}{2}}$$ $$ = \bigg(\frac{2m + 1}{2}\bigg)\bigg(\frac{2m - 1}{2}\bigg)\bigg(\frac{2m - 3}{2}\bigg)......\frac{3}{2}\frac{1}{2}\sqrt{\pi}$$ $$ = \frac{(2m + 1)(2m - 1)(2m - 3)....31*\sqrt{\pi}}{2^{m+1}}$$

Taking equation (B), multiplying and dividing by $\frac{2}{x}$

$J_{\frac{1}{2}}(x) = \sqrt{\frac{x}{2}}* \frac{2}{x} \sum\limits_{m = 0}^{\infty} \frac{(-1)^m x^{2m + 1}}{2^{2m + 1} m! \sqrt{m + \frac{3}{2}}}......(C)$

$2^{2m + 1}m! = 2^{2m + 1}2^m m(m - 1).....4.1\\ 2^{2m + 1} m ! = 2^{2m + 1}(2m)(2m - 2)(2m - 4)....4.2$

Now

$2^{2m + 1}m! \sqrt{m + \frac{3}{2}} = \frac{2^{2m + 1} (2m)(2m - 2)(2m - 4)...4.2)[(2m+1)(2m - 1)(2m - 3).....3*1 \sqrt{\pi}]}{2^{2m + 1}}\\ = (2m + 1)(2m)(2m - 1)(2m - 2)(2m - 3)....4.3.2.1\sqrt{\pi}$

$2^{2m + 1} m! \sqrt{m + \frac{3}{2}} = (2m + 1)! \sqrt{\pi}$

Now, from equation (C)

$J_{\frac{1}{2}}(x) = \sqrt{\frac{2}{x}} \sum\limits_{m = 0}^{\infty}\frac{(-1)^m x^{2m + 1}}{(2m +1)! \sqrt{\pi}}$

$J_{\frac{1}{2}}(x) = \sqrt{\frac{2}{\pi x}} \sum\limits_{m = 0}^{\infty}\frac{(-1)^m x^{2m + 1}}{(2m +1)! }$

$J_{\frac{1}{2}}(x) = \sqrt{\frac{2}{\pi x}}sinx$

Now,

$\frac{d}{dx}x^n J_{n}(x) = x^n J_{n - 1}(x)$

Put, n = $\frac{1}{2}$

$\therefore \frac{d}{dx}\sqrt{x} J_{\frac{1}{2}}(x) = \sqrt{x}J_{\frac{-1}{2}}(x)\\ \therefore \frac{d}{dx}\sqrt{x}\sqrt{\frac{2}{\pi x}}sinx = \sqrt{x}J_{\frac{-1}{2}}(x)\\ \therefore \sqrt{\frac{2}{\pi}} \frac{d}{dx}sinx = \sqrt{x}J_{\frac{-1}{2}}(x)$

$\therefore J_{\frac{-1}{2}}(x) = \sqrt{\frac{2}{\pi x}}cosx$

Now substituting $J_{\frac{1}{2}}(x)$ & $J_{\frac{-1}{2}}(x)$ in equation (A)

$\therefore J_{\frac{3}{2}}(x) = \frac{1}{x}\sqrt{\frac{2}{\pi x}}sinx - \sqrt{\frac{2}{\pi x}}cosx \\ = \sqrt{\frac{2}{\pi x}}\bigg(\frac{sinx}{x} - cosx \bigg)$

$\therefore J_{\frac{3}{2}}(x) \sqrt{\frac{2}{\pi x}}\bigg(\frac{sinx}{x} - cosx \bigg)$