Thread: Trigonometric Differentiation

hi,

here y = 3xsinx

dy/dx = 3sinx + 3xcosx

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

Originally Posted by mortoleeuk

hi,

here y = 3xsinx

dy/dx = 3sinx + 3xcosx

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

Use the product rule (uv)' = u'v + uv', with u = 3x and v = sin(x) such that uv = 3xsin(x).
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).