Question

1. - / 5.55 points

differentiate.
$f(t) = \frac{2t}{2 + \sqrt{t}}$
$f(t) = \square$

Explanation:

Step1: Identify numerator and denominator

Let \( u = 2t \), \( v = 2 + \sqrt{t} = 2 + t^{1/2} \)

Step2: Compute derivatives of u and v

\( u' = 2 \), \( v' = \frac{1}{2}t^{-1/2} = \frac{1}{2\sqrt{t}} \)

Step3: Apply quotient rule \( f'(t)=\frac{u'v - uv'}{v^2} \)

Substitute values: \( f'(t)=\frac{2(2+\sqrt{t}) - 2t(\frac{1}{2\sqrt{t}})}{(2+\sqrt{t})^2} \)

Step4: Simplify numerator

\( 4 + 2\sqrt{t} - \sqrt{t} = 4 + \sqrt{t} \)

Step5: Write final derivative

\( f'(t)=\frac{4 + \sqrt{t}}{(2+\sqrt{t})^2} \)

Answer:

\(\frac{4 + \sqrt{t}}{(2 + \sqrt{t})^2}\)