Question

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

Answer 1

`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`