In solving the problem, it is determined that the derivative of h(x) = sin^2x + cosx.
0 is less than x which is less than 2pi
was: h'(x) = cosxsinx +sinxcosx - sinx which then simplified to 2sinxcosx-sinx which then simplified to sinx(2cosx-1)
When do the derivative, I get cos^2x -sinx
Would you explain what I am doing wrong in finding the derivative if you can figure that out, and would you please provide a step by step analysis of how the you got the derivative above.© BrainMass Inc. brainmass.com December 15, 2022, 4:58 pm ad1c9bdddf
Your mistake is, when you do derivative for sin^2x, you get ((sinx)')^2=(cosx)^2=cos^2x. You simply exchange the order of square and sin.
The solution assists with showing the steps to finding the derivative.