How do you find the first order derivative of sin(x) and cos(x) functions?
The general rules are:d(sin x)/ dx = cos xd(cos x)/ dx = -sinxHowever, if you have a constant k in the function such as y = sin(kx), the derivative will be dy/dx = kcos(kx). (For y = cos(kx), the derivative is dy/dx = -ksin(kx)).Now this formula doesn't work for all cases, e.g. those that have k = x2 , for that we use these formulas:If y = sin(f(x)), the derivative is dy/dx = f'(x)cos(f(x))and if y = cos(f(x)), the derivative is dy/dx = -f'(x)sin(f(x))If these formulas look confusing, note that the f(x) has just replaced the k to show that the k can have any order. This formula just means that the drerivative of the function f(x) (denoted by f'(x)) inside the sin or cos goes infront (is multiplied to it). Here are some examples:What is the derivative of y = sin(3x)? here, f(x) = 3x, therefore f'(x) = 3 as per normal differentiation rules therefore, dy/dx = 3cos(3x)What is the derivative of y = cos(2x3)? here, f(x) = 2x3, therefore f'(x) = 6x2 therefore, dy/dx = -6x2sin(2x3)
