Question

425.Посчитайте $f'(a)$ при:
a)$f(x)=\frac{2+x}{3x-1},a=-1$
c)$f(x)=3x*sin(x),a=\frac{\pi}{6}$
e)$f(x)=\frac{ln(x)}{2x},a=1$

Answer 1

$1)\ \ f(x)=\dfrac{2+x}{3x-1}\ \ ,\ \ a=-1\\\\\\f'(x)=\dfrac{1\cdot (3x-1)-3\cdot (2+x)}{(3x-1)^2}=\dfrac{3x-1-6-3x}{(3x-1)^2}=\dfrac{-7}{(3x-1)^2}\\\\\\f'(-1)=\dfrac{-7}{(-3-1)^2}=-\dfrac{7}{16}$

$2)\ \ f(x)=3x\cdot sinx\ \ ,\ \ a=\dfrac{\pi}{6}\\\\\\f'(x)=3\cdot sinx+3x\cdot cosx\\\\\\f'(\dfrac{\pi}{6})=3\cdot sin\dfrac{\pi}{6}+3\cdot \dfrac{\pi}{6}\cdot cos\dfrac{\pi}{6}=3\cdot \dfrac{1}{2}+\dfrac{\pi}{2}\cdot \dfrac{\sqrt3}{2}=\dfrac{6+\pi \sqrt3}{4}$

$3)\ \ f(x)=\dfrac{lnx}{2x}\ \ ,\ \ a=1\\\\\\y'=\dfrac{\dfrac{1}{x}\cdot 2x-2\cdot lnx}{4x^2}=\dfrac{2\cdot (1-lnx)}{4x^2}\\\\\\a=1\ \ ,\ \ f'(1)=\dfrac{2\cdot (1-ln1)}{4}=\dfrac{2}{4}=\dfrac{1}{2}$