Using the power rule:

$\displaystyle y=0.0002{t}^{ 3} -0.4{t}^{ 2}+4+\frac{100}{{t}^{2 } } $

$\displaystyle y=0.0006{t}^{ 2}-0.8t-200{t}^{-3 } $

just trying to check my answer

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- May 24th 2011, 06:05 PMpurplec16Find the derivative
Using the power rule:

$\displaystyle y=0.0002{t}^{ 3} -0.4{t}^{ 2}+4+\frac{100}{{t}^{2 } } $

$\displaystyle y=0.0006{t}^{ 2}-0.8t-200{t}^{-3 } $

just trying to check my answer - May 24th 2011, 06:21 PMSudharaka
- May 25th 2011, 04:28 AMpurplec16
Ok...thank you

- May 25th 2011, 04:48 AMAckbeet
Although, technically, you should write

$\displaystyle \dot{y}$ or $\displaystyle y'$

for your derivative. - May 25th 2011, 05:38 AMpurplec16
i did i just forgot to put it on the forum my apologies.

- May 25th 2011, 05:41 AMAckbeet
No problem!