Basic Derivatives, Level 1
y = 3
y' = 0
y = sin(x) [csc(x) + sec(x)]
y' = sec2(x)
(distribute first)
y = 3x2 sin x
y' = 3x2 cos x + 6x sin x
y = (5x - 9)4
y' = 20(5x - 9)3
f(x) = 5x2-2x+1,000,000
f'(x) = 10x - 2
f(x) = (2-x3) / x2
f'(x) = -4x-3 - 1
Write f as 2x-2 -x
y = (5x - 2)/(x2 + 1)
(and simplify the numerator)
y' = (-5x2 + 4x + 5)/(x2 + 1)2
f(x) = sin 2x
2 cos 2x
f(x) = -3x - cot(x)
f'(x) = -3 + csc2(x)
y = ln (ex)
y' = 1
y=x because the functions are inverses
f(x) = -2(3x - 2x2)sec(x)
f'(x) = -2(3-4x)sec(x) -2(3x - 2x2)sec(x)tan(x)
y = csc(5x-1)
y' = -5csc(5x-1)cot(5x-1)
f(x) = 3sinx - 2cosx
3cosx + 2sinx
f(x)=4sin-1(x)
f'(x) = 4 / (1-x2)1/2
f(x) = (2x + 5)/(x)1/2
f'(x) = (2x1/2 - (1/2)(2x + 5)x-1/2)/x
multiply by 2x1/2 / 2x1/2
= (2x - 5)/(2x3/2)
y = x2(3x+5)10
y' = 2x(3x+5)10 + 30x2(3x+5)9
y = 2tan(x) - 5sec(x)
y' = 2sec2(x) - 5sec(x)tan(x)
y = sin2(3x)+cos2(3x)
y' = 0 (because y = 1 by the Pythagorean Identity)
y = x3sin(x)cos(x)
y' = 3x2sin(x)cos(x) + x3cos2(x) - x3sin2(x)
f(x) = cos3(4x)
f'(x) = -12sin(4x)cos2(4x)