Question

$Calcule f'(x), sendof(t)= \frac{ \sqrt{t-1} }{ \sqrt{t+1} } =0$

Answer 1

$f(t)=\dfrac{\sqrt{t-1}}{\sqrt{t+1}}$

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Queremos calcular a derivada de $f$ em relação a $t.$

( usando a regra do quociente e a regra da cadeia )

$f'(t)=\left(\dfrac{\sqrt{t-1}}{\sqrt{t+1}} \right )'\\\\\\ =\dfrac{\big(\sqrt{t-1}\big)'\cdot\sqrt{t+1}-\sqrt{t-1}\cdot \big(\sqrt{t+1}\big)'}{\big(\sqrt{t+1}\big)^2}\\\\\\ =\dfrac{\left(\frac{1}{2\sqrt{t-1}}\cdot (t-1)' \right )\cdot\sqrt{t+1}-\sqrt{t-1}\cdot \left(\frac{1}{2\sqrt{t+1}}\cdot (t+1)' \right )}{\big(\sqrt{t+1}\big)^2}\\\\\\ =\dfrac{\left(\frac{1}{2\sqrt{t-1}}\cdot 1 \right )\cdot\sqrt{t+1}-\sqrt{t-1}\cdot \left(\frac{1}{2\sqrt{t+1}}\cdot 1 \right )}{\big(\sqrt{t+1}\big)^2}$

$\therefore~~\boxed{\begin{array}{c}f'(t)=\dfrac{\frac{1}{2\sqrt{t-1}}\cdot\sqrt{t+1}-\sqrt{t-1}\cdot \frac{1}{2\sqrt{t+1}}}{\big(\sqrt{t+1}\big)^2} \end{array}}$

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Caso queira, pode tentar simplificar ainda mais a função derivada:

$f'(t)=\dfrac{\frac{1}{2\sqrt{t-1}}\cdot\sqrt{t+1}-\sqrt{t-1}\cdot \frac{1}{2\sqrt{t+1}}}{\big(\sqrt{t+1}\big)^2}\cdot \dfrac{2\,\sqrt{t-1}\cdot \sqrt{t+1}}{2\,\sqrt{t-1}\cdot \sqrt{t+1}}\\\\\\ =\dfrac{\left[\frac{1}{2\sqrt{t-1}}\cdot\sqrt{t+1}-\sqrt{t-1}\cdot \frac{1}{2\sqrt{t+1}} \right ]\cdot 2\,\sqrt{t-1}\cdot \sqrt{t+1}}{\big(\sqrt{t+1}\big)^2\cdot 2\,\sqrt{t-1}\cdot \sqrt{t+1}}\\\\\\ =\dfrac{\big(\sqrt{t+1}\big)^2-\big(\sqrt{t-1}\big)^2}{\big(\sqrt{t+1}\big)^2\cdot 2\,\sqrt{t-1}\cdot \sqrt{t+1}}\\\\\\ =\dfrac{t+1-(t-1)}{(t+1)^2\cdot 2\,\sqrt{t-1}\cdot \sqrt{t+1}}\\\\\\ =\dfrac{\big(\sqrt{t+1}\big)^2-\big(\sqrt{t-1}\big)^2}{\big(\sqrt{t+1}\big)^2\cdot 2\,\sqrt{t-1}\cdot \sqrt{t+1}}\\\\\\ =\dfrac{\diagup\!\!\!\! 2}{(t+1)^{3/2}\cdot \diagup\!\!\!\! 2\,\sqrt{t-1}}\\\\\\\\ \therefore~~\boxed{\begin{array}{c}f'(t)=\dfrac{1}{(t+1)^{3/2}\cdot \sqrt{t-1}} \end{array}}$