Single Variable Calculus, Chapter 3, 3.5, Section 3.5, Problem 36

Determine the derivative of the function $\displaystyle f(t) = \sqrt{\frac{t}{t^2+4}}$


$
\begin{equation}
\begin{aligned}
f'(t) &= \frac{d}{dt} \left(\frac{t}{t^2+4} \right)^{\frac{1}{2}}\\
\\
f'(t) &= \frac{1}{2} \left(\frac{t}{t^2+4} \right)^{\frac{-1}{2}} \frac{d}{dt} \left(\frac{t}{t^2+4} \right)\\
\\
f'(t) &= \frac{1}{2} \left(\frac{t}{t^2+4} \right)^{\frac{-1}{2}} \left[ \frac{(t^2+4)\frac{d}{dt} (t) - (t) \frac{d}{dt} (t^2+4)}{(t^2+4)^2}\right]\\
\\
f'(t) &= \frac{1}{2} \left(\frac{t}{t^2+4} \right)^{\frac{-1}{2}} \left[ \frac{(t^2+4)(1) - (t)(2t)}{(t^2+4)^2} \right]\\
\\
f'(t) &= \frac{1}{2} \left(\frac{t}{t^2+4} \right)^{\frac{-1}{2}} \left[ \frac{t^2 + 4 - 2t^2}{(t^2+4)^2} \right]\\
\\
f'(t) &= \frac{1}{2} \left(\frac{t}{t^2+4} \right)^{\frac{-1}{2}} \left[ \frac{-t^2+4}{(t^2+4)^2} \right]\\
\\
f'(t) &= \frac{1}{2} \frac{(t)^{\frac{-1}{2}}(4-t^2)}{(t^2+4)^{\frac{-1}{2}}(t^2+4)^2}\\
\\
f'(t) &= \frac{4-t^2}{2(t)^{\frac{1}{2}}(t^2+4)^{\frac{3}{2}}}
\end{aligned}
\end{equation}
$