To find the derivative of the function \(y = r \cdot e^{-x} \cdot \cos^2(x(x^2+x+1))\), you can use logarithmic differentiation. Here's the process:
1. Take the natural logarithm (ln) of both sides of the equation:
\(\ln(y) = \ln(r \cdot e^{-x} \cdot \cos^2(x(x^2+x+1)))\)
2. Use logarithm properties to simplify the expression:
\(\ln(y) = \ln(r) + \ln(e^{-x}) + \ln(\cos^2(x(x^2+x+1)))\)
3. Apply the properties of logarithms:
\(\ln(y) = \ln(r) - x + 2\ln(\cos(x(x^2+x+1)))\)
4. Differentiate both sides with respect to \(x\):
\(\frac{1}{y} \cdot \frac{dy}{dx} = -1 + 2\ln(\cos(x(x^2+x+1))) \cdot \frac{d}{dx}(x^3+x^2+x)\)
5. Now, you can solve for \(\frac{dy}{dx}\), the derivative of \(y\):
\(\frac{dy}{dx} = y \cdot \left[2\ln(\cos(x(x^2+x+1))) \cdot \frac{d}{dx}(x^3+x^2+x) - 1\right]\)
You can further simplify the expression by computing the derivative of \(x^3+x^2+x\) and then substitute it back in.