Calculus: Early Transcendentals, Chapter 3, 3.6, Section 3.6, Problem 40

y=(e^-x*(cos(x))^2)/(x^2+x+1)

Taking the natural logarithm of both sides and applying the properties of logarithms, we get

logy=loge^-x+2logcosx -log(x^2+x+1)
logy=-x+2logcosx-log(x^2+x+1)

Differentiating both sides with respect to x, we get

1/y dy/(dx)=-1+(2/cosx)(-sinx) -(1/(x^2+x+1))(2x+1)
1/y dy/(dx)=-1-2tanx-(2x+1)/(x^2+x+1)
dy/dx=y(-1-2tanx-(2x+1)/(x^2+x+1))
dy/dx=((e^-x(cos(x)))^2/(x^2+x+1))(-1-2tanx-(2x+1)/(x^2+x+1))
dy/dx=-((e^-x(cos(x))^2)/(x^2+x+1))(((x^2+x+1+2(x^2+x+1)(tanx)+2x+1))/(x^2+x+1))
dy/dx=-((e^-x(cos(x))^2)(x^2+3x+2+2(x^2+x+1)tanx))/(x^2+x+1)^2