Question

$Derivada f(t)= (3t^{2})/(t-1) +(5t)/( t-1)- (1)/(t-1)$

Answer 1

$f(t)= \dfrac{3t^2}{t-1} +\dfrac{5t}{t-1} -\dfrac{1}{t-1}\\\\\\f(t)=\dfrac{3t^2+5t-1}{t-1} \\\\\\Regra\,\,do\,\,quociente\\\\f'(t)=\dfrac{f'(t).g(t)-f(t).g'(t)}{[g(t)]^2}}\\\\\\f(t)=3t^2+5t-1\,\,\,\,\,\,\,\,\,\,g(t)=t-1\\f'(t)=6t+5 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,g'(t) = 1\\\\\\f'(t)= \dfrac{(6t+5).(t-1)-[(3t^2+5t-1).1]}{(t-1)^2}\\\\\\f'(t)= \dfrac{6t^2-6t+5t-5-(3t^2+5t-1)}{(t-1)^2}\\\\\\\boxed{f'(t)=\dfrac{3t^2-6t-4}{(t-1)^2}}$

Answer 2

f(t) é uma soma de frações homogêneas

           $f(t) = \frac{3t^2+5t-1}{t-1} \\ \\ \frac{d}{dt}( \frac{3t^2+5t-1}{t-1}) \\ \\ = \frac{(t-1) \frac{d}{t}(3t^2+5t-1)-(3t^2+5t-1) \frac{d}{dt}((t-1) }{(t-1)^2} \\ \\ = \frac{(t-1)(6t+5)-(3t^2+5t-1)(1)}{t^2-2t+1} \\ \\ = \frac{6t^2+5t-6t-5-3t^2-5t+1}{t^2-2t+1} \\ \\ = \frac{3t^2-6t-4}{t^2-2t+1}$