Question

Va rog functiile derivate si explicit sa inteleg..

Answer 1

1)

$f'(x)=\frac{(x^2+1)'*2x-(x^2+1)*(2x)'}{(2x)^2}=\frac{2x*2x-2(x^2+1)}{4x^2}=\frac{4x^2-2x^2-2}{4x^2}=\frac{2x^2-2}{4x^2}=\frac{2(x^2-1)}{4x^2}=\frac{x^2-1}{2x^2}$

2)

$f'(x)=[(x+1)^3-x^3]'=[(x+1)^3]'-(x^3)'=3(x+1)^{3-1}(x+1)'-3x^2=3(x+1)^2-3x^2=3(x^2+2x+1)-3x^2=3x^2+6x+3-3x^2=6x+3$

3)

$f'(x)=\frac{(x^2-x+7)'(x+1)-(x^2-x+7)(x+1)'}{(x+1)^2}=\frac{(2x-1)(x+1)-(x^2-x+7)}{(x+1)^2}=\frac{2x^2+2x-x-1-x^2+x-7}{(x+1)^2}=\frac{x^2+2x-8}{(x+1)^2}$