Монгол Бодлогын Сан

Заавар: $$[f(g(x))]^\prime=f^\prime(g(x))\cdot g^\prime(x)$$ $$\left(\dfrac{u(x)}{v(x)}\right)^\prime=\dfrac{u^\prime(x)v(x)-u(x)\cdot v^\prime(x)}{v^2(x)}$$

Бодолт: $(\ln x)^\prime=\dfrac{1}{x}$ тул $$y^\prime=\Big(\ln \sqrt{\displaystyle\frac{1+\sin x}{1-\sin x}}\Big)^\prime=\dfrac12\Big(\ln \displaystyle\frac{1+\sin x}{1-\sin x}\Big)^\prime=$$ $$=\dfrac12\cdot\dfrac{1}{\frac{1+\sin x}{1-\sin x}}\cdot\Big(\dfrac{1+\sin x}{1-\sin x}\Big)^\prime=$$ $$\dfrac12\cdot\dfrac{1}{\frac{1+\sin x}{1-\sin x}}\cdot\dfrac{\cos x(1-\sin x)-(-\cos x)(1+\sin x)}{(1-\sin x)^2}=$$ $$=\dfrac{2\cos x}{2(1+\sin x)(1-\sin x)}=\dfrac{\cos x}{1-\sin^2x}=\dfrac{1}{\cos x}$$ тул $$y^\prime\Big(\dfrac{\pi}{4}\Big)=\dfrac{1}{\cos\frac{\pi}{4}}=\dfrac{1}{\frac{\sqrt2}{2}}=\sqrt2$$ байна.