Question

Найдите значение производной в точке x0:
а) f(x)=4x^2+6x+3, x0=1;
б)f(x)=x/1+x^2 ,x0=0;
в)f(x)=(3x^2+1)(3x^2-1) ,x0=1;
г)f(x)=2x*cosx, x0=Pi/4

Answer 1

$f(x)=4x^2+6x+3 \ f'(x)=8x+6 \ f'(x_0)=f'(1)=8*1+6=14 \ \ f(x)= frac{x}{1+x^2} \ f'(x)= frac{1*(1+x^2)-x*2x}{(1+x^2)^2}= frac{1+x^2-2x^2}{(1+x^2)^2}= frac{1-x^2}{(1+x^2)^2} \f'(0)=frac{1-0^2}{(1+0^2)^2}= frac{1}{1}=1 \ \ f(x)=(3x^2+1)(3x^2-1)=(3x^2)^2-1^2=9x^4-1 \ f'(x)=9*4x^3=36x^3\ f'(1)=36*1^3=36$
$f(x)=2x*cosx \ f'(x)=2*cosx+2x*(-sinx)=2cosx-2xsinx \ f'( frac{ pi }{4})=2cosfrac{ pi }{4}-2*frac{ pi }{4}*sinfrac{ pi }{4}=2*frac{sqrt{2}}{2}-2* frac{ pi }{4}*frac{sqrt{2}}{2}= sqrt{2}(1- frac{pi}{4})$