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

[latex]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 [/latex] [latex]f(x)=2x*cosx \\ f'(x)=2*cosx+2x*(-sinx)=2cosx-2xsinx \\ f'( \frac{ \pi }{4})=2cos\frac{ \pi }{4}-2*\frac{ \pi }{4}*sin\frac{ \pi }{4}=2*\frac{\sqrt{2}}{2}-2* \frac{ \pi }{4}*\frac{\sqrt{2}}{2}= \sqrt{2}(1- \frac{\pi}{4}) [/latex]