find the derivative of y with respect to θ if y = ln(\\frac{e^{11θ}}{1 + e^{11θ}}).\n\\frac{dy}{dθ}=□

Explanation:

Step1: Use the quotient - rule for logarithms

Recall that $\ln(\frac{a}{b})=\ln(a)-\ln(b)$. So, $y = \ln(e^{11\theta})-\ln(1 + e^{11\theta})$. Since $\ln(e^{11\theta}) = 11\theta$, then $y=11\theta-\ln(1 + e^{11\theta})$.

Step2: Differentiate term - by - term

The derivative of $11\theta$ with respect to $\theta$ is $11$ (using the power rule $\frac{d}{d\theta}(k\theta)=k$ where $k = 11$). For the second term, let $u = 1+e^{11\theta}$. Then $\frac{du}{d\theta}=11e^{11\theta}$ (using the chain - rule, derivative of $e^{ax}$ is $ae^{ax}$). The derivative of $\ln(u)$ with respect to $\theta$ is $\frac{1}{u}\cdot\frac{du}{d\theta}$ by the chain - rule. Substituting $u = 1 + e^{11\theta}$ and $\frac{du}{d\theta}=11e^{11\theta}$, we get $\frac{11e^{11\theta}}{1 + e^{11\theta}}$.

Step3: Calculate the derivative of $y$

$\frac{dy}{d\theta}=\frac{d}{d\theta}(11\theta)-\frac{d}{d\theta}\ln(1 + e^{11\theta})$. $\frac{dy}{d\theta}=11-\frac{11e^{11\theta}}{1 + e^{11\theta}}$. We can simplify this as follows: [ \begin{align*} \frac{dy}{d\theta}&=\frac{11(1 + e^{11\theta})-11e^{11\theta}}{1 + e^{11\theta}}\ &=\frac{11+11e^{11\theta}-11e^{11\theta}}{1 + e^{11\theta}}\ &=\frac{11}{1 + e^{11\theta}} \end{align*} ]

Answer:

$\frac{11}{1 + e^{11\theta}}$