Chat with our AI personalities
f(x) = √(2x -3) f(x) = (2x - 3)^(1/2) f'(x) = (1/2)[(2x - 3)^(1/2 - 1)](2) f'(x) = (2x - 3)^(-1/2) f'(x) = 1/[(2x - 3)^(1/2)] f'(x) = 1/√(2x -3)
x2/(3x2 - 5x - 2) - [2x/ (3x + 1)][1/(x - 2)] = x2/(3x2 - 6x + x - 2) - 2x/(3x + 1)(x - 2) = x2/[3x(x - 2) + (x - 2)] - 2x/(3x + 1)(x - 2) = x2/(3x + 1)(x - 2) - 2x/(3x + 1)(x - 2) = (x2 - 2x)/(3x + 1)(x - 2) = x(x - 2)/(3x + 1)(x - 2) = x/(3x + 1)
tan x + (tan x)(sec 2x) = tan 2x work dependently on the left sidetan x + (tan x)(sec 2x); factor out tan x= tan x(1 + sec 2x); sec 2x = 1/cos 2x= tan x(1 + 1/cos 2x); LCD = cos 2x= tan x[cos 2x + 1)/cos 2x]; tan x = sin x/cos x and cos 2x = 1 - 2 sin2 x= (sin x/cos x)[(1 - 2sin2 x + 1)/cos 2x]= (sin x/cos x)[2(1 - sin2 x)/cos 2x]; 1 - sin2 x = cos2 x= (sin x/cos x)[2cos2 x)/cos 2x]; simplify cos x= (2sin x cos x)/cos 2x; 2 sinx cos x = sin 2x= sin 2x/cos 2x= tan 2x
(1+cosx)(1-cosx)= 1 +cosx - cosx -cos^2x (where ^2 means squared) = 1-cos^2x = sin^2x (sin squared x)
3x2 + 2x + 2 = 0 Can not be factored. You can however solve it for x: 3x2 + 2x = -2 x2 + 2x/3 = -2/3 x2 + 2x/3 + 1/9 = -2/3 + 1/9 (x + 1/9)2 = -5/9 x + 1/9 = ±√(-5/9) x = -1/9 ± i√(5/9) x = -1/9 ± i√5 / 3 x = -1/9 ± 3i√5 / 9 x = (-1 ± 3i√5) / 9