We are asked to find the derivative of the function $y = \frac{e^{3x}}{(x^2+3)^2(1+\cos x)^4}$ using logarithmic differentiation.

1. Problem Description

We are asked to find the derivative of the function $y = \frac{e^{3x}}{(x^2+3)^2(1+\cos x)^4}$ using logarithmic differentiation.

2. Solution Steps

Let $y = \frac{e^{3x}}{(x^2+3)^2(1+\cos x)^4}$.

Taking the natural logarithm of both sides, we have

$\ln y = \ln \left(\frac{e^{3x}}{(x^2+3)^2(1+\cos x)^4}\right)$

Using the logarithm properties $\ln(\frac{a}{b}) = \ln a - \ln b$ and $\ln(ab) = \ln a + \ln b$, we can rewrite the equation as:

$\ln y = \ln(e^{3x}) - \ln((x^2+3)^2(1+\cos x)^4)$

$\ln y = \ln(e^{3x}) - [\ln(x^2+3)^2 + \ln(1+\cos x)^4]$

$\ln y = \ln(e^{3x}) - \ln(x^2+3)^2 - \ln(1+\cos x)^4$

Using the logarithm property $\ln(a^b) = b\ln a$, we have

$\ln y = 3x\ln e - 2\ln(x^2+3) - 4\ln(1+\cos x)$

Since $\ln e = 1$, we have

$\ln y = 3x - 2\ln(x^2+3) - 4\ln(1+\cos x)$

Now, differentiate both sides with respect to $x$:

$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} [3x - 2\ln(x^2+3) - 4\ln(1+\cos x)]$

$\frac{1}{y} \frac{dy}{dx} = 3 - 2 \frac{1}{x^2+3} (2x) - 4 \frac{1}{1+\cos x} (-\sin x)$

$\frac{1}{y} \frac{dy}{dx} = 3 - \frac{4x}{x^2+3} + \frac{4\sin x}{1+\cos x}$

Multiply both sides by $y$:

$\frac{dy}{dx} = y \left(3 - \frac{4x}{x^2+3} + \frac{4\sin x}{1+\cos x}\right)$

Substitute $y = \frac{e^{3x}}{(x^2+3)^2(1+\cos x)^4}$ back into the equation:

$\frac{dy}{dx} = \frac{e^{3x}}{(x^2+3)^2(1+\cos x)^4} \left(3 - \frac{4x}{x^2+3} + \frac{4\sin x}{1+\cos x}\right)$

3. Final Answer

$\frac{dy}{dx} = \frac{e^{3x}}{(x^2+3)^2(1+\cos x)^4} \left(3 - \frac{4x}{x^2+3} + \frac{4\sin x}{1+\cos x}\right)$