Question

derivada de:
a) 2t/2+√t

Answer 1

$f(t)=\dfrac{2t}{2+\sqrt{t}}=\dfrac{a(t)}{b(t)}$, onde $a(t)=2t$ e $b(t)=2+\sqrt{t}$

Derivando $f$ pela regra do quociente:

$f'(t)=\dfrac{a'(t)b(t)-a(t)b'(t)}{[b(t)]^{2}}\\\\\\f'(t)=\dfrac{(2+\sqrt{t})\frac{d}{dt}(2t)-2t\frac{d}{dt}(2+\sqrt{t})}{(2+\sqrt{t})^{2}}\\\\\\f'(t)=\dfrac{(2+\sqrt{t})\cdot2-2t\cdot(0+\frac{1}{2\sqrt{t}})}{(2+\sqrt{t})^{2}}\\\\\\f'(t)=\dfrac{4+2\sqrt{t}-\frac{t}{\sqrt{t}}}{(2+\sqrt{t})^{2}}$

Note que $\dfrac{t}{\sqrt{t}}=\dfrac{t\sqrt{t}}{\sqrt{t}\sqrt{t}}=\dfrac{t\sqrt{t}}{t}=\sqrt{t}~~~para~t~\textgreater~0$

Então:

$f'(t)=\dfrac{4+2\sqrt{t}-\sqrt{t}}{(2+\sqrt{t})^{2}}\\\\\\\boxed{\boxed{f'(t)=\dfrac{4+\sqrt{t}}{(2+\sqrt{2})^{2}}}}$