Use logarithmic differentiation to find the derivative of the function. show steps please!
y=(e^-xcos^2(x))/(x^2+x+1)
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.
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!
To find the derivative of the function y = (e^(-x) * cos^2(x))/(x^2 + x + 1), we will use logarithmic differentiation. Here's how you can do it step by step:
Step 1: Take the natural logarithm of both sides of the equation to simplify the expression and make logarithmic differentiation possible:
ln(y) = ln[(e^(-x) * cos^2(x))/(x^2 + x + 1)]
Step 2: Simplify the right-hand side of the equation using logarithmic properties:
ln(y) = ln(e^(-x) * cos^2(x)) - ln(x^2 + x + 1)
Step 3: Expand the logarithm using the properties of logarithms:
ln(y) = -x + ln(cos^2(x)) - ln(x^2 + x + 1)
Step 4: Differentiate both sides of the equation implicitly with respect to x:
(d/dx) ln(y) = (d/dx) [-x + ln(cos^2(x)) - ln(x^2 + x + 1)]
Step 5: Use the chain rule and derivative rules to find the derivatives on the right-hand side:
(d/dx) ln(y) = -1 + (2cos(x) * -sin(x))/cos^2(x) - (1/(x^2 + x + 1)) * (2x + 1)
Step 6: Note that the derivative of ln(y) with respect to x is (1/y) * (dy/dx) by the chain rule. Substitute y back in for ln(y):
(1/y) * (dy/dx) = -1 + (2cos(x) * -sin(x))/cos^2(x) - (1/(x^2 + x + 1)) * (2x + 1)
Step 7: Multiply both sides of the equation by y to isolate the dy/dx term:
dy/dx = y * [-1 + (2cos(x) * -sin(x))/cos^2(x) - (1/(x^2 + x + 1)) * (2x + 1)]
Step 8: Rewrite y in terms of the original function:
dy/dx = [(e^(-x) * cos^2(x))/(x^2 + x + 1)] * [-1 + (2cos(x) * -sin(x))/cos^2(x) - (1/(x^2 + x + 1)) * (2x + 1)]
Simplifying this expression gives you the derivative of the original function using logarithmic differentiation.