find the derivative of y with respect to θ if y = ln(\\frac{e^{12θ}}{1 + e^{12θ}}).\\frac{dy}{dθ}=□

Explanation:

Step1: Use logarithm properties

$y=\ln(e^{12\theta})-\ln(1 + e^{12\theta})=12\theta-\ln(1 + e^{12\theta})$

Step2: Differentiate term - by - term

The derivative of $12\theta$ with respect to $\theta$ is $12$. For the second term, let $u = 1+e^{12\theta}$, then $\frac{du}{d\theta}=12e^{12\theta}$. By the chain - rule, the derivative of $\ln(u)$ with respect to $\theta$ is $\frac{1}{u}\cdot\frac{du}{d\theta}$. Substituting $u = 1 + e^{12\theta}$ and $\frac{du}{d\theta}=12e^{12\theta}$, we get $\frac{12e^{12\theta}}{1 + e^{12\theta}}$.

Step3: Calculate the derivative of $y$

$\frac{dy}{d\theta}=12-\frac{12e^{12\theta}}{1 + e^{12\theta}}=\frac{12(1 + e^{12\theta})-12e^{12\theta}}{1 + e^{12\theta}}=\frac{12+12e^{12\theta}-12e^{12\theta}}{1 + e^{12\theta}}=\frac{12}{1 + e^{12\theta}}$

Answer:

$\frac{12}{1 + e^{12\theta}}$