Question

Составьте уравнение производной к графику функции f в точке с абсциссой x0, если:

Answer 1

$1)f(x)=x^{3} +x\\\\f'(x)=(x^{3} )'+x'=3x^{2} +1\\\\f'(-1)=3*(-1)^{2}+1=3+1=\boxed4\\\\\\2)f(x)=x^{2} -\sqrt{x}\\\\f'(x)=(x^{2})'-(\sqrt{x})'=2x-\frac{1}{2\sqrt{x} } \\\\f'(4)=2*4-\frac{1}{2*\sqrt{4} }=8-\frac{1}{4}=8-0,25=\boxed{7,75}$

$3)f(x)=\frac{x+2}{x-4}\\\\f'(x)=(\frac{x+2}{x-4} )'=\frac{(x+2)'*(x-4)-(x+2)*(x-4)'}{(x-4)^{2} }=\frac{x-4-x-2}{(x-4)^{2} }=-\frac{6}{(x-4)^{2} } \\\\f'(3)=-\frac{6}{(3-4)^{2} }=\boxed{-6} \\\\\\4)f(x)=Sinx\\\\f'(x)=(Sinx)'=Cosx\\\\f'(-\frac{\pi }{3})=Cos(-\frac{\pi }{3})=Cos\frac{\pi }{3}=\frac{1}{2} =\boxed{0,5}$