Question

Найти похідну функції

Answer 1

Ответ:

$900)\ \ y=\dfrac{1+sin2x}{1-sin2x}\\\\y'=\dfrac{2cos2x(1-sin2x)-(1+sin2x)(-2cos2x)}{(1-sin2x)^2}=\\\\\\=\dfrac{2\, cos2x-2\, cos2x\cdot sin2x+2cos2x+2cos2x\cdot sin2x}{(1-sin2x)^2}=\dfrac{4\, cos2x}{(1-sin2x)^2}$

$902)\ \ r=cos^2\Big(\dfrac{\pi}{4}-\dfrac{\varphi }{2}\Big)\\\\\\r'=2\, cos\Big(\dfrac{\pi}{4}-\dfrac{\varphi }{2}\Big)\cdot sin\Big(\dfrac{\pi}{4}-\dfrac{\varphi }{2}\Big)\cdot \dfrac{1}{2}=\dfrac{1}{2}\cdot sin\Big(\dfrac{\pi}{2}-\varphi \Big)$

$904)\ \ f(t)=\sqrt{1+cos^2t^2}\\\\f'(t)=\dfrac{1}{2\sqrt{1+cos^2t^2}}\cdot 2cost^2\cdot (-sint^2)\cdot 2t=-\dfrac{2t\cdot cost^2\cdot sint^2}{\sqrt{1+cos^2t^2}}=\\\\\\=-\dfrac{t\cdot sin(2t^2)}{\sqrt{1+cos^2t^2}}$