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

1 Answer

Mar 3, 2017

#dy/dt = 24sqrtt -2t#

Explanation:

#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#

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