Question

Обчисліть похідну в точці х0:

Answer 1

$\displaystyle\bf\\1)\\\\f(x)=2x^{3}-8\sqrt{x} +\sqrt{6} \\\\\\f'(x)=2\cdot(x^{3} )'-8\cdot(\sqrt{x} )'+(\sqrt{6} )'=2\cdot 3x^{2} -8\cdot \frac{1}{2\sqrt{x} } +0=\\\\\\=6x^{2} -\frac{4}{\sqrt{x} } \\\\\\f'(x_{0} )=f'(1)=6\cdot 1^{2} -\frac{4}{\sqrt{1} } =6-4=2\\\\\\\boxed{f'(1)=2}\\\\\\2)\\\\f(x)=8\cdot x^{-3} \\\\\\f'(x)=8\cdot(x^{-3} )'=8\cdot(-3x^{-4} )=-\frac{24}{x^{4} }\\\\\\f'(x_{0} )=f'(2)=-\frac{24}{2^{4} }=-\frac{24}{16} =-1,5\\\\\\\boxed{f'(2)=-1,5}$

Answer 2

Ответ:

Объяснение:

$\boldsymbol{(x^n )' = n\cdot x^{n-1}} \\\\ \boldsymbol{(\sqrt{x})' = \cfrac{1}{2\sqrt{x} } }$

$\hspace{-1,1em}a) ~ f(x) = 2x^3 - 8 \sqrt{x} + \sqrt{6 } ~,~ x_0 = 1 \\\\ f'(x) = 6x^2 - \dfrac{8}{2\sqrt{x} } = 6x^2 - \dfrac{4}{\sqrt{x} } \\\\ \boxed{ f'(1) = 6 -4 = 2}$

$\hspace{-1em }b) ~ f(x) = 8 x^{-3} ~ , ~ x_0 \\\\ f'(x) = 8 \cdot (-3x^{-4} ) =- \dfrac{24}{x^4} \\\\ \boxed{f(2 ) = -\cfrac{24}{2^4 } = -1,5 }$