Question

Qual a derivada de √cos5x?

Answer 1

y = (sen5x + cos5x)/(sen5x - cos5x)

y ' = [(5cos5x - 5sen5x)*(sen5x - cos5x) - (sen5x + cos5x)*(5cos5x + 5sen5x)]/(sen5x - cos5x)²

y ' = [- 5(sen5x - cos5x)*(sen5x - cos5x) - 5(sen5x + cos5x)*(cos5x + sen5x)]/(sen5x - cos5x)²

y ' = - 5[(sen5x - cos5x)*(sen5x - cos5x) +(sen5x + cos5x)*(cos5x + sen5x)]/(sen5x - cos5x)²

y ' = - 5[(sen5x - cos5x)² + (sen5x + cos5x)²]/(sen5x - cos5x)²

y ' = - 5[(sen²5x - 2sen5xcos5x + cos²5x + sen²5x + 2sen5xcos5x + cos²5x]/(sen5x - cos5x)²

y ' = - 5[(sen²5x + cos²5x) + (sen²5x + cos²5x)]/(sen5x - cos5x)²

y ' = - 5[(1) + (1)]/(sen5x - cos5x)²

y ' = - 10/(sen5x - cos5x)²