Question

Derivar

$y= \frac{1}{3}ln(t+1)- \frac{1}{6} ln(t^2-t+1)+ \frac{1}{ \sqrt{3} }arctg( \frac{2t-1}{ \sqrt{3} } )$



R: $y'= \frac{1}{t^3+1}$

Answer 1

 Olá!
 
 Derivando a função termo a termo, afim de facilitar a visualização, temos:

Termo I:

$\\ \mathsf{y_1 = \frac{1}{3} \cdot \ln (t + 1)} \\\\ \mathsf{y_1' = \frac{1}{3} \cdot \frac{1}{t + 1} \cdot 1} \\\\ \boxed{\mathsf{y_1' = \frac{1}{3(t + 1)}}}$

Termo II:

$\\ \mathsf{y_2 = \frac{1}{6} \cdot \ln (t^2 - t + 1)} \\\\ \mathsf{y_2' = \frac{1}{6} \cdot \frac{1}{t^2 - t + 1} \cdot (2t - 1)} \\\\ \boxed{\mathsf{y_2' = \frac{2t - 1}{6(t^2 - t + 1)}}}$
 
 Quando ao Termo III, temos que: $\mathsf{f(x) = \arctan x \Rightarrow f'(x) = \frac{x'}{1 + x^2}}$
 
 Isto posto,

$\\ \mathsf{y_3 = \frac{1}{\sqrt{3}} \cdot \arctan \left ( \frac{2t - 1}{\sqrt{3}} \right )} \\\\\\ \mathsf{y_3' = \frac{1}{\sqrt{3}} \cdot \frac{\frac{2 \cdot \sqrt{3} - (2t - 1) \cdot 0}{3}}{1 + \frac{4t^2 - 4t + 1}{3}}} \\\\\\ \mathsf{y_3' = \frac{1}{\sqrt{3}} \cdot \frac{2\sqrt{3}}{3} \div \frac{4t^2 - 4t + 4}{3}} \\\\\\ \mathsf{y_3' = \frac{1}{1} \cdot \frac{2}{3} \div \frac{3}{4(t^2 - t + 1)}} \\\\\\ \mathsf{y_3' = \frac{1}{2(t^2 - t + 1)}}$
 
 Com efeito,

$\\ \mathsf{y' = \frac{1}{3(t + 1)} - \frac{2t - 1}{6(t^2 - t + 1)} + \frac{1}{2(t^2 - t + 1)}} \\\\\\ \mathsf{y' = \frac{1}{3(t + 1)/2(t^2 - t + 1)} - \frac{2t - 1}{6(t^2 - t + 1)/(t + 1)} + \frac{1}{2(t^2 - t + 1)/3(t + 1)}} \\\\\\ \mathsf{y' = \frac{2(t^2 - t + 1) - (2t - 1)(t + 1) + 3(t + 1)}{6(t + 1)(t^2 - t + 1)}} \\\\\\ \mathsf{y' = \frac{2t^2 - 2t + 2 - (2t^2 + 2t - t - 1) + 3t + 3}{6(t^3 + 1)}}$

$\\ \mathsf{y' = \frac{6}{6(t^3 + 1)}} \\\\\\ \boxed{\boxed{\mathsf{\mathsf{y' = \frac{1}{t^3 + 1}}}}}$