Asked by Tim
Use logarithmic differentiation to find the derivative of the function. show steps please!
y=(e^-xcos^2(x))/(x^2+x+1)
y=(e^-xcos^2(x))/(x^2+x+1)
Answers
Answered by
Steve
log y = log (e^-x - xcos^2(x)) - log(x^2+x+1)
1/y y' = (-e^-x - (cos^2 x - 2xcosx*sinx))/(e^x - xcos^2 x) - (2x+1)/(x^2+x+1)
y' = (-e^x - cos^2 x + xsin2x) - (2x+1)/(x^2+x+1)^2
If you want to put it all over the common denominator, it gets fairly messy on top.
1/y y' = (-e^-x - (cos^2 x - 2xcosx*sinx))/(e^x - xcos^2 x) - (2x+1)/(x^2+x+1)
y' = (-e^x - cos^2 x + xsin2x) - (2x+1)/(x^2+x+1)^2
If you want to put it all over the common denominator, it gets fairly messy on top.
Answered by
Steve
actually, I see I did forget to include a factor of 1/(x^2+x+1) in the first term. Dang!
Luckily, it's not my job to fix it!
Luckily, it's not my job to fix it!
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