$F\left(x\right)\={\left(\frac{x-\mathrm{\π}}{x^2-4}\right)}^7$$\stackrel{\text{assign as function}}{\to }$$F$

$F\prime \left(x\right)$ = $\frac{7{\left(x-\mathrm{\π}\right)}^6}{{\left(x^2-4\right)}^7}-\frac{14{\left(x-\mathrm{\π}\right)}^{7}x}{{\left(x^2-4\right)}^8}$$\stackrel{\text{simplify}}{\=}$$\frac{7{\left(-x\+\mathrm{\π}\right)}^6\left(-x^2-4\+2x\mathrm{\π}\right)}{{\left(x^2-4\right)}^8}$

Stepwise solution

Define $f\left(u\right)\=u^7$ as the "outer" function, and $g\left(x\right)\=\frac{x-\pi }{x^2-4}$ as the "inner" function, so that $f\left(x\right)\=f\left(g\left(x\right)\right)$. The Chain rule applies first, but the derivative of $g$, the "inner" function, requires the Quotient rule.