می‌دانیم: ${f}'(\frac{\pi }{8})g(\frac{\pi }{8})+{g}'(\frac{\pi }{8})(\frac{\pi }{8})=(f(x)g(x){)}'(\frac{\pi }{8})$  پس داریم: $f(x)g(x)={{(\cos x-\frac{1}{\cos x})}^{10}}{{(\sin x-\frac{1}{\sin x})}^{10}}={{(\frac{{{\cos }^{2}}x-1}{\cos x})}^{10}}{{(\frac{{{\sin }^{2}}x-1}{\sin x})}^{10}}$ $={{(\frac{-{{\sin }^{2}}x}{\cos x}\times \frac{-{{\cos }^{2}}x}{\sin x})}^{10}}={{(\sin x\cos x)}^{10}}={{(\frac{1}{2}\sin 2x)}^{10}}=\frac{1}{{{2}^{10}}}{{\sin }^{10}}2x$  $(f(x)g(x){)}'=\frac{10}{{{2}^{10}}}\times 2{{\sin }^{9}}2x\cos 2x$  $(f(x)g(x){)}')(\frac{\pi }{8})=\frac{10}{{{2}^{9}}}\times {{(\frac{\sqrt{2}}{2})}^{9}}(\frac{\sqrt{2}}{2})=\frac{10}{{{2}^{14}}}$