Expert: Sherry Wallin - 9/25/2009

Question
Differentiate the function f(x)=2cos^5 3x+xlnx

Answer
Let u = 3x so you have 2cos^5u +xlnx
Let's take the derivative of each term since the derivative of a sum is the sum of the derivatives.
(2cos^5u)' = 2(cos u)^5 = 5*2cos^4u(-sin^5u)u' = -3*10cos^4 3x *sin^5 3x, now do:
(xlnx)' = x(1/x) + 1*lnx = 1 + lnx
So the overall answer is: -30cos^4 3x *sin^5 3x + 1 + lnx

Math Prof

x*lnx is the product of two functions x and lnx. To find the derivative use the product rule:
the first function 'x' times the derivative of the second function '1/x' plus the second function 'lnx' times the derivative of the first function '1': x(1/x) + lnx(1) = 1 + lnx