Question

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Номер 8.8 и 8.10

Answer 1

Ответ:

$\displaystyle 8.8)\ \ 1)\ f(x)=ln(6x-5)\ \ ,\ \ f'(x)=\dfrac{6}{6x-5}\ \ ,\\\\f'(3)=\dfrac{6}{18-5}=\dfrac{6}{13}\\\\\\2)\ f(x)=8\, ln\frac{x}{2}\ \ ,\ \ f'(x)=8\cdot \frac{2}{x}\cdot \frac{1}{2}=\frac{8}{x}\ \ ,\ \ \ f'(\frac{1}{2})=16\\\\\\3)\ \ f(x)=lg(x^2-5x+8)\ \ ,\ \ f'(x)=\frac{2x-5}{x^2-5x+8}\ \ ,\ \ f'(2)=\frac{-1}{2}=-\frac{1}{2}\\\\\\4)\ \ f(x)=lncos\frac{x}{3}\ \ ,\ \ f'(x)=\frac{-1}{3cos\frac{x}{3}}\cdot sin\frac{x}{3}=-\frac{1}{3}\, tg\dfrac{x}{3}\ \ ,$

$f'(\dfrac{\pi }{2})=-\dfrac{1}{3}\cdot tg\dfrac{\pi}{6}=-\dfrac{\sqrt3}{9}$

$8.10)\ \ 1)\ \ f(x)=e^{1-x}\ \ ,\ \ k=f'(x_0)\\\\f'(x)=-e^{1-x}\ \ ,\ \ k=f'(1)=-e^0=-1\\\\2)\ \ f(x)=log_5(x+2)\ \,\ \ x_0=-1\\\\f'(x)=\dfrac{1}{(x+2)\, ln5}\ \ ,\ \ k=f'(-1)=\dfrac{1}{ln5}$