Monday

April 21, 2014

April 21, 2014

Posted by **Tom** on Sunday, June 17, 2012 at 12:44pm.

- Calculus -
**Reiny**, Sunday, June 17, 2012 at 1:07pmwork it as a chain-rule question, by treating it like

y = (tanx)^3

dy/dx = 3(tanx)^2 (sec^2 x)

= 3(tan^2 x)(sec^2 x)

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