Question

Найдите значение производной функции y=g(x) в точке x0:
1) g(x)=√x-9x², x0=1
2) g(x)=8√x+1/x, x0=16

Answer 1

$1)\; \; g(x)=\sqrt{x}-9x^2\; \; ,\; \; x_0=1\\\\g'(x)= \frac{1}{2\sqrt{x}} -18x\\\\g'(1)= \frac{1}{2} -18=-17\frac{1}{2}=-17,5\\\\2)\; \; g(x)= \frac{8\sqrt{x}+1}{x}\; \; ,\; \; x_0=16\\\\g'(x)=\frac{\frac{8}{2\sqrt{x}}\cdot x-(8\sqrt{x}+1)}{x^2} =\frac{4\sqrt{x}-8\sqrt{x}-1}{x^2}=\frac{-4\sqrt{x}-1}{x^2} \\\\g'(16)=\frac{-4\cdot 4-1}{16^2} = -\frac{17}{256}$