Question

Estou com dúvida nestas duas questões, alguém poderia mim ajudar, retribuirei com 99 pontos.

Answer 1

Questão 1:

$f(x)=\dfrac{x}{x^2-3x+2}$


Calcularemos a derivada de $f$ pela regra do quociente:

$f'(x)=\dfrac{(x)'\cdot (x^2-3x+2)-x\cdot (x^2-3x+2)'}{(x^2-3x+2)^2}\\\\\\ f'(x)=\dfrac{1\cdot (x^2-3x+2)-x\cdot (2x-3)}{(x^2-3x+2)^2}\\\\\\ f'(x)=\dfrac{x^2-3x+2-2x^2+3x}{(x^2-3x+2)^2}\\\\\\ \boxed{\begin{array}{c} f'(x)=\dfrac{2-x^2}{(x^2-3x+2)^2} \end{array}}$


Portanto,

$f'\left(\frac{1}{2} \right )=\dfrac{2-\left(\frac{1}{2}\right )^2}{\left[\left(\frac{1}{2}\right )^2-3\cdot \frac{1}{2}+2\right]^2}\\\\\\ =\dfrac{2-\frac{1}{4}}{\left[\frac{1}{4}-\frac{3}{2}+2\right ]^2}\\\\\\ =\dfrac{\frac{8}{4}-\frac{1}{4}}{\left[\frac{1}{4}-\frac{6}{4}+\frac{8}{4}\right ]^2}\\\\\\ =\dfrac{\frac{7}{4}}{\left[\frac{3}{4}\right]^2}\\\\\\ =\dfrac{\left(\frac{7}{4}\right )}{\left(\frac{9}{16}\right )}\\\\\\ =\dfrac{7}{4}\cdot \dfrac{16}{9}\\\\\\ =\dfrac{28}{9}$

Resposta: alternativa C.

______________________

Questão 2:

$f(t)=\dfrac{1}{t^3}+\dfrac{t^4}{2}-\dfrac{4}{t^2}\\\\\\ f(t)=t^{-3}+\dfrac{1}{2}\,t^4-4t^{-2}$


Derivando pela regra da potência:

$f'(t)=\left(t^{-3}+\dfrac{1}{2}\,t^4-4t^{-2} \right )'\\\\\\ f'(t)=-3t^{-3-1}+\dfrac{1}{2}\cdot 4t^{4-1}-4\cdot (-2)t^{-2-1}\\\\\\ f'(t)=-3t^{-4}+\dfrac{1}{2}\cdot 4t^{3}-4\cdot (-2)t^{-3}\\\\\\ f'(t)=-3t^{-4}+2t^{3}+8t^{-3}\\\\\\ \boxed{\begin{array}{c}f'(t)=-\dfrac{3}{t^4}+2t^{3}+\dfrac{8}{t^3} \end{array}}$


Resposta: alternativa A.