Упр.247 ГДЗ Колягин Ткачёва 11 класс (Алгебра)
2) корень x+4/4x;
3) x/корень x+3;
4) sinx+cosx/sinx-cosx.
$$y=\frac{1-\cos 2x}{1+\cos 2x}$$
По правилу производной частного:
$$ y'(x)=\frac{(1-\cos 2x)'(1+\cos 2x)-(1-\cos 2x)(1+\cos 2x)’}{(1+\cos 2x)^2} $$
$$ y'(x)=\frac{2\sin 2x(1+\cos 2x)+2\sin 2x(1-\cos 2x)}{(1+\cos 2x)^2} $$
$$ y'(x)=\frac{4\sin 2x}{(1+\cos 2x)^2} $$
$$y=\frac{\sqrt{x+4}}{4x}$$
$$ y'(x)=\frac{\left(\sqrt{x+4}\right)’ \cdot 4x-\sqrt{x+4}\cdot (4x)’}{(4x)^2} $$
$$ y'(x)=\frac{\frac{1}{2\sqrt{x+4}}\cdot 4x-\sqrt{x+4}\cdot 4}{16x^2} $$
$$ y'(x)=\frac{2x-4(x+4)}{16x^2\sqrt{x+4}}=-\frac{x+8}{8x^2\sqrt{x+4}} $$
$$y=\frac{x}{\sqrt{x+3}}$$
$$ y'(x)=\frac{(\,x\,)’\sqrt{x+3}-x\left(\sqrt{x+3}\right)’}{x+3} $$
$$ y'(x)=\frac{\sqrt{x+3}-x\cdot \frac{1}{2\sqrt{x+3}}}{x+3} =\frac{2(x+3)-x}{2\sqrt{x+3}\,(x+3)} $$
$$ y'(x)=\frac{x+6}{2(x+3)\sqrt{x+3}}=\frac{x+6}{2(x+3)^{3/2}} $$
$$y=\frac{\sin x+\cos x}{\sin x-\cos x}$$
$$ y'(x)=\frac{(\cos x-\sin x)(\sin x-\cos x)-(\sin x+\cos x)(\cos x+\sin x)}{(\sin x-\cos x)^2} $$
$$ y'(x)=\frac{-(\sin x-\cos x)^2-(\sin x+\cos x)^2}{(\sin x-\cos x)^2} $$
$$ y'(x)=\frac{-2(\sin^2 x+\cos^2 x)}{(\sin x-\cos x)^2} =-\frac{2}{(\sin x-\cos x)^2} $$
Ответ
$$ 1)\; y’=\frac{4\sin 2x}{(1+\cos 2x)^2};\quad 2)\; y’=-\frac{x+8}{8x^2\sqrt{x+4}}; $$
$$ 3)\; y’=\frac{x+6}{2(x+3)^{3/2}};\quad 4)\; y’=-\frac{2}{(\sin x-\cos x)^2}. $$
