Question

Determine o valor da derivada primeira de f(x)=x/(x^2-3x+2) no ponto x+1/2.

Answer 1

$f(x)=\frac{x}{x^2-3x+2} \\ \\ f'(x)= \frac{(x^2-3x+2).x'-x.(x^2-3x+2)'}{(x^2-3x+2)^2} \\ \\ f'(x)= \frac{(x^2-3x+2).1-x(2x-3)}{(x^2-3x+2)^2} \\ \\ f('x)= \frac{x^2-3x+2-2x^2+3x}{(x^2-3x+2)^2} \\ \\ f'(x)= \frac{2}{(x^2-3x+2)^2}$

Calcular a derivada em x = 1/2, consiste em calcular f '(1/2)

$f'(\frac{1}{2} = \frac{2}{[ \frac{1}{2} )^2-3. \frac{1}{2}+2]^2 } \\ \\ f'( \frac{1}{2}) = \frac{2}{( \frac{1}{4} - \frac{3}{2}+2)^2 } \\ \\ f'( \frac{1}{2} )= \frac{2}{( \frac{1-6+8}{4})^2 } \\ \\ f'( \frac{1}{2} )= \frac{2}{ (\frac{3}{4})^2 } \\ \\ f'( \frac{1}{2}) = \frac{2}{ \frac{9}{16} } \\ \\ f'( \frac{1}{2})= \frac{32}{9}$