What is the first differential of $y = t^(3/2)(16-sqrtt)$ ?

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Answer 1 · 2017-03-03T11:54:11

$dy/dt = 24sqrtt -2t$

$y = t^(3/2)(16-sqrtt)$

= $16t^(3/2) - t^(1/2+3/2)$

$= 16t^(3/2) - t^2$

Applying the power rule to both terms:

$dy/dt = 16* 3/2t^(1/2) - 2t$

$= 24sqrtt-2t$

What is the first differential of y = t^(3/2)(16-sqrtt)y = t^(3/2)(16-sqrtt)?