Question

Номера - 22, 40, 36
Помогите пж

Answer 1

Пошаговое объяснение:

$f(g(x))' = g'(x) f'(g(x))$

$f^n(x)' =n * f^{n-1}(x) * f'(x)$

22

$\displaystyle f(\theta) = \sin\sqrt\theta\\f(\theta)'=(\sin\sqrt\theta)' = (\sqrt\theta)'\cos\sqrt\theta= \cfrac{\cos\sqrt\theta}{2\sqrt{\theta}}$

36

$\displaystyle y = \sqrt{\cfrac{1+\cos x}{1- \cos x}}\\y' = \left(\sqrt{\cfrac{1+\cos x}{1- \cos x}}\right)' = \left(\cfrac{1+\cos x}{1- \cos x}\right)' * \cfrac{1}{2\sqrt{\cfrac{1+\cos x}{1- \cos x}}}$

$\left(\cfrac{1+\cos x}{1- \cos x}\right)' = \cfrac{(1+\cos x)'(1-\cos x) - (1 + \cos x)(1 - \cos x)'}{(1-\cos x)^2} =\\= \cfrac{-\sin x + \sin x \cos x - \sin x - \sin x \cos x}{(1 - \cos x)^2} = \cfrac{-2\sin x}{1 - 2\cos x + \cos^2 x}$

$\displaystyle y' = \left(\cfrac{1+\cos x}{1- \cos x}\right)' * \cfrac{1}{2\sqrt{\cfrac{1+\cos x}{1- \cos x}}} =-\cfrac{\sin x(\sqrt{1-\cos x})}{(1 - 2\cos x + \cos^2 x)(\sqrt{1+\cos x})}$

40

$\displaystyle y = \cfrac{1}{3} \ tg^3 \cfrac{1}{x}\\y' = \cfrac{1}{3} \left(3 \ tg^2 \cfrac{1}{x} \ \left(tg \cfrac{1}{x}\right)' \ \right)\\\left(tg \cfrac{1}{x}\right)' \ \right) = -\cfrac{1}{x^2} * \cfrac{1}{\cos^2 \cfrac{1}{x}}\\y' =- \cfrac{tg^2 \cfrac{1}{x}}{x^2\cos^2 \cfrac{1}{x}}$