Question

найдите производную (похідну), пожалуйста​

Answer 1

Ответ:

$1)\ \ y=cos^2\Big(\dfrac{\pi}{4}-5x\Big)\ \ ,\ \ \ \ \ \ (u^2)'=2u\cdot u'\ ,\ \ u=cos(\frac{\pi}{4}-5x)\ \ ,\ \ \Big(\dfrac{1}{u}\Big)'=-\dfrac{u'}{u^2}\\\\y'=2\cdot cos\Big(\dfrac{\pi}{4}-5x\Big)\cdot \Big(-sin\Big(\dfrac{\pi}{4}-5x\Big)\Big)\cdot (-5)=5\cdot sin\Big(\dfrac{\pi}{2}-10x\Big)=5\cdot cos10x\\\\\\y=\dfrac{5}{cos^2\Big(\dfrac{\pi}{4}-5x\Big)}\ \ ,\ \ y'=-\dfrac{25\, cos10x}{cos^4\Big(\dfrac{\pi}{4}-5x\Big)}$

$2)\ \ y=sin^2\Big(\dfrac{\pi}{6}-2x\Big)\ \ ,\ \ \ \ \ \ (u^2)'=2u\cdot u'\ ,\ \ u=sin(\frac{\pi}{6}-2x)\\\\y'=2\cdot sin\Big(\dfrac{\pi}{6}-2x\Big)\cdot cos\Big(\dfrac{\pi}{6}-2x\Big)\cdot (-2)=-2\cdot sin\Big(\dfrac{\pi}{3}-4x\Big)\\\\\\y=\dfrac{4}{sin^2\Big(\dfrac{\pi}{6}-2x\Big)}\ \ ,\ \ \ y'=-\dfrac{-2\cdot sin\Big(\dfrac{\pi}{3}-4x\Big)}{sin^4\Big(\dfrac{\pi}{6}-2x\Big)}=\dfrac{2\cdot sin\Big(\dfrac{\pi}{3}-4x\Big)}{sin^4\Big(\dfrac{\pi}{6}-2x\Big)}$

$3)\ \ y=5e^{5x+1}\ \ ,\ \ \ y'=5e^{5x+1}\cdot 5=25\, e^{5x+1}\\\\\\4)\ \ y=6\cdot 3^{2-3x}\ \ ,\ \ \ y'=6\cdot e^{2-3x}\cdot ln3\cdot (-3)=-18\cdot ln3\cdot e^{2-3x}$

$5)\ \ y=\dfrac{1}{5-3x}\ \ ,\ \ \ \Big(\dfrac{1}{u}\Big)'=\dfrac{-u'}{u^2}\\\\\\y'=\dfrac{-(-3)}{(5-3x)^2}=\dfrac{3}{(5-3x)^2}$