hi,

here y = 3xsinx

dy/dx = 3sinx + 3xcosx

could someone run through the process of getting that, exam 2moro...Thanx

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- May 22nd 2005, 09:15 AMmortoleeukTrigonometric Differentiation
hi,

here y = 3xsinx

dy/dx = 3sinx + 3xcosx

could someone run through the process of getting that, exam 2moro...Thanx - May 22nd 2005, 10:52 AMhpeQuote:

Originally Posted by**mortoleeuk**

Then u' = 3 and v' = cos(x).

Thus u'v = 3sin(x) and uv' = 3xcos(x).

Thus dy/dx = (uv)' = u'v + uv' = 3sin(x) + 3xcos(x).