>>593

(√f)/g　の微分　ただし、f=3x^2-4x+4、g=x^2-x+1

((√f)/g)'
={(√f)’*g-(√f)*g’}/g^2
={(1/2)(1/√f)*f’*g-(√f)*g’}/g^2
={(1/2)*f’*g-f*g’}/{g^2*√f}
={(1/2)*(6x-4)*(x^2-x+1)-(3x^2-4x+4)*(2x-1)}/{(x^2-x+1)^2*√(3x^2-4x+4)}
={(3x-2)*(x^2-x+1)-(3x^2-4x+4)*(2x-1)}/{(x^2-x+1)^2*√(3x^2-4x+4)}
={(3x^3-5x^2+5x-2)-(6x^3-11x^2+12x-4)}/{(x^2-x+1)^2*√(3x^2-4x+4)}
={-3x^3+6x^2-7x+2}/{(x^2-x+1)^2*√(3x^2-4x+4)}