y=[√(x+2)](3-x)^4/(x+1)^5求dy/dx

Question

y=[√(x+2)](3-x)^4/(x+1)^5 求dy/dx

Answer 1

lny=(1/2)ln(x+2)+4ln(3-x)-5ln(x+1)
(1/y)dy/dx=1/[2(x+2)]-4/(3-x)-5/(x+1)
dy/dx=y{1/[2(x+2)]-4/(3-x)-5/(x+1)}
dy/dx=(√(x+2)](3-x)^4/(x+1)^5) {1/[2(x+2)]-4/(3-x)-5/(x+1)}