Wednesday

May 6, 2015

May 6, 2015

Posted by **sean** on Wednesday, May 14, 2008 at 12:12am.

- maths -
**drwls**, Wednesday, May 14, 2008 at 2:19amUse the chain rule separately for each of the two terms. The first term can be differentiated as follows:

Let u = tan 3x and v = u^2

tan^2(3x) = v{u(x)}

d/dx [tan^2(3x)] = (dv/du)*(du/dx)

= 2u * 3 sec^2 3x

= 6 tan(3x)*sec^2(3x)

Use a similar procedure to differentiate sec^2(3x)