use logarithmic diff. to find the derivative of the function. Show steps please! so I can see how it is done. Thank you so much!
y=(e^(-x)cos^(2)(x))/(x^(2)+x+1)
y = (e^-x) (cosx)^2 (x^2 + x + 1)
take ln of both sides
ln y = ln e^-x + ln (cosx)^2 + ln(x^2 + x + 1)
= -x + 2 ln(cosx) + ln(x^2 + x + 1)
now differentiate
y' / y = -1 + 2(-sinx/cosx) + (2x+1)/(x^2 + x + 1)
= -1 - 2tanx + (2x+1)/(x^2 + x + 1)
y' = y(-1 - 2tanx + (2x+1)/(x^2 + x + 1))
= [ (e^-x) (cosx)^2 (x^2 + x + 1) ] * [ -1 - 2tanx + (2x+1)/(x^2 + x + 1) ]
sure hope they don't expect us to simplify this
log y = log (e^-x) + log cos^2(x) - log(x^2+x+1)
log y = -x + 2log cos x - log(x^2+x+1)
1/y y' = -1 - 2tanx - (2x+1)/(x^2+x+1)
y' = -(1 + 2tanx + (2x+1)/(x^2+x+1)) * (e^-x cos^2x)/(x^2+x+1)
Now, you can massage that for a few more steps, to get something that pleases you
To find the derivative of the function using logarithmic differentiation, follow these steps:
Step 1: Take the natural logarithm of both sides of the equation.
ln(y) = ln[(e^(-x)cos^2(x))/(x^2+x+1)]
Step 2: Use the properties of logarithms to simplify the expression.
ln(y) = ln(e^(-x)cos^2(x)) - ln(x^2+x+1)
ln(y) = -x + 2ln(cos(x)) - ln(x^2+x+1)
Step 3: Differentiate both sides of the equation implicitly with respect to x.
(1/y) * y' = -1 + 2 * (1/cos(x)) * (-sin(x)) - [(2x+1)/(x^2+x+1)]
Step 4: Simplify and solve for y'.
y' = y * [2sin(x)/cos(x) - (2x+1)/(x^2+x+1)]
y' = (e^(-x)cos^2(x))/(x^2+x+1) * [2sin(x)/cos(x) - (2x+1)/(x^2+x+1)]
Therefore, the derivative of the function y = (e^(-x)cos^2(x))/(x^2+x+1) with logarithmic differentiation is:
y' = (e^(-x)cos^2(x))/(x^2+x+1) * [2sin(x)/cos(x) - (2x+1)/(x^2+x+1)]
Note: The steps shown above involve the application of logarithmic differentiation, which is a technique used to differentiate a function when it contains a combination of exponential, trigonometric, and polynomial terms.