Знайдіть похідну f(x)=(3х−1)√2х−2 х0=3

$f(x)=(3x-1) \sqrt{2x-2}$

$f'(x)=[(3x-1) \sqrt{2x-2}]'=$

$=(3x-1)'* \sqrt{2x-2}+(3x-1)*( \sqrt{2x-2} )'=$

$=[(3x)'-(1)']* \sqrt{2x-2}+(3x-1)*[(2x-2})^{ \frac{1}{2} }]' =$

$=[3*(x)'-0]* \sqrt{2x-2}+(3x-1)*[\frac{1}{2}*(2x-2})^{ \frac{1}{2}-1 }*(2x-2)']=$

$=[3*1-0]* \sqrt{2x-2}+(3x-1)*\frac{1}{2}*(2x-2})^{ -\frac{1}{2}}*(2x-2)'=$

$=3 \sqrt{2x-2}+(3x-1)* \frac{1}{2}* \frac{1}{(2x-2)^ \frac{1}{2} }}*[(2x)'-(2)']=$

$=3 \sqrt{2x-2}+(3x-1)* \frac{1}{2 \sqrt{2x-2}}}*[2*(x)'-0]=$

$=3 \sqrt{2x-2}+(3x-1)* \frac{1}{2 \sqrt{2x-2}}}*[2*1-0]=$

$=3 \sqrt{2x-2}+(3x-1)* \frac{2*1}{2* \sqrt{2x-2}}}=$

$=3 \sqrt{2x-2}+(3x-1)* \frac{1}{\sqrt{2x-2}}} =3 \sqrt{2x-2}+\frac{3x-1}{\sqrt{2x-2}}}$

$f'(x_0)=f'(3)=3 \sqrt{2*3-2}+\frac{3*3-1}{\sqrt{2*3-2}}}=3 \sqrt{4} + \frac{8}{ \sqrt{4} } =$

$=3*2+ \frac{8}{2}=6+4=10$

Відповідь: $10$