2. Ha#TH y' Hy'' 2.26. ) x=sqrt [3]((t-1)^2) y=sqrt (t-1)

To find the derivatives of \( x \) and \( y \) with respect to \( t \), we need to differentiate each equation separately.<br /><br />Given:<br />\[ x = \sqrt[3]{(t-1)^2} \]<br />\[ y = \sqrt{t-1} \]<br /><br />First, let's find \( \frac{dx}{dt} \):<br /><br />\[ x = (t-1)^{2/3} \]<br /><br />Using the chain rule:<br />\[ \frac{dx}{dt} = \frac{d}{dt} (t-1)^{2/3} \]<br />\[ \frac{dx}{dt} = \frac{2}{3} (t-1)^{-1/3} \cdot \frac{d}{dt} (t-1) \]<br />\[ \frac{dx}{dt} = \frac{2}{3} (t-1)^{-1/3} \cdot 1 \]<br />\[ \frac{dx}{dt} = \frac{2}{3} (t-1)^{-1/3} \]<br />\[ \frac{dx}{dt} = \frac{2}{3} \cdot \frac{1}{(t-1)^{1/3}} \]<br />\[ \frac{dx}{dt} = \frac{2}{3(t-1)^{1/3}} \]<br /><br />Next, let's find \( \frac{dy}{dt} \):<br /><br />\[ y = (t-1)^{1/2} \]<br /><br />Using the chain rule:<br />\[ \frac{dy}{dt} = \frac{d}{dt} (t-1)^{1/2} \]<br />\[ \frac{dy}{dt} = \frac{1}{2} (t-1)^{-1/2} \cdot \frac{d}{dt} (t-1) \]<br />\[ \frac{dy}{dt} = \frac{1}{2} (t-1)^{-1/2} \cdot 1 \]<br />\[ \frac{dy}{dt} = \frac{1}{2} (t-1)^{-1/2} \]<br />\[ \frac{dy}{dt} = \frac{1}{2} \cdot \frac{1}{(t-1)^{1/2}} \]<br />\[ \frac{dy}{dt} = \frac{1}{2(t-1)^{1/2}} \]<br /><br />So, the derivatives are:<br />\[ \frac{dx}{dt} = \frac{2}{3(t-1)^{1/3}} \]<br />\[ \frac{dy}{dt} = \frac{1}{2(t-1)^{1/2}} \]