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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Calculando a derivada:

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


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