d the derivative of the function y=2t^4-16sqrt (t)+(6)/(t) )=square

To find the derivative of the function \( y = 2t^4 - 16\sqrt{t} + \frac{6}{t} \), we will differentiate each term separately.<br /><br />1. Differentiate \( 2t^4 \):<br /> \[<br /> \frac{d}{dt}(2t^4) = 2 \cdot 4t^3 = 8t^3<br /> \]<br /><br />2. Differentiate \( -16\sqrt{t} \):<br /> Recall that \( \sqrt{t} = t^{1/2} \). Using the power rule:<br /> \[<br /> \frac{d}{dt}(-16t^{1/2}) = -16 \cdot \frac{1}{2}t^{-1/2} = -8t^{-1/2} = -\frac{8}{\sqrt{t}}<br /> \]<br /><br />3. Differentiate \( \frac{6}{t} \):<br /> Recall that \( \frac{6}{t} = 6t^{-1} \). Using the power rule:<br /> \[<br /> \frac{d}{dt}(6t^{-1}) = 6 \cdot (-1)t^{-2} = -6t^{-2} = -\frac{6}{t^2}<br /> \]<br /><br />Now, combine the results:<br />\[<br />\frac{dy}{dt} = 8t^3 - \frac{8}{\sqrt{t}} - \frac{6}{t^2}<br />\]<br /><br />So, the derivative of the function \( y = 2t^4 - 16\sqrt{t} + \frac{6}{t} \) is:<br />\[<br />\boxed{8t^3 - \frac{8}{\sqrt{t}} - \frac{6}{t^2}}<br />\]