Question

Найти производную функции y = f(x) в точке x=a, когда:


а) $f(x)=\sqrt{2x+3},x=-1$


б) $f(x)=cos(\frac{x}{2}),x=-\frac{\pi }{2}$

Answer 1

а) f’(x) = ( (2x+3)^(1/2) )’ = 1/2(2x+3)^(-1/2) * 2 = (2x+3)^(1/2) = 1/(sqrt(2x+3))

f’(a) = 1/(sqrt(2*(-1) + 3)) = 1/1 = 1.

б) f’(x) = -sin (x/2) * 1/2 = -sin(x/2)/2.

f’(a) = -sin(-pi/4) * 1/2 = sin(pi/4) * 1/2 = (sqrt(2))/4.

Ответ: a) 1; б) (sqrt(2))/4.

Answer 2

$1)f(x)=\sqrt{2x+3} \\\\f'(x)=\frac{1}{2\sqrt{2x+3}}*(2x+3)'=\frac{2}{2\sqrt{2x+3} }=\frac{1}{\sqrt{2x+3}}\\\\f'(-1)=\frac{1}{\sqrt{2*(-1)+3}}=\frac{1}{\sqrt{-2+3} }=1\\\\\boxed{f'(-1)=1}\\\\\\2)f(x)=Cos\frac{x}{2}\\\\f'(x)=-Sin\frac{x}{2}*(\frac{x}{2})'=-\frac{1}{2} Sin\frac{x}{2}\\\\f'(-\frac{\pi }{2})=-\frac{1}{2}Sin\frac{-\frac{\pi }{2}}{2} =-\frac{1}{2}Sin(-\frac{\pi }{4})=\frac{1}{2}Sin\frac{\pi }{4}=\frac{1}{2}*\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{4}$

$\boxed{f'(-\frac{\pi }{2})=\frac{\sqrt{2}}{4}}$