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

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Chapter 3, 3.6 - Problem 40 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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`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`

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