# `y = (e^-x (cos(x))^2)/(x^2 + x +1)` Use logarithmic differentiation to find the derivative of the function.

*print*Print*list*Cite

Expert Answers

gsarora17 | Certified Educator

`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`