Question

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

Explanation:

Step1: Use log - property

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

Step2: Simplify $\ln(e^{12\theta})$

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

Step3: 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 $\frac{du}{d\theta}=12e^{12\theta}$.
The derivative of $\ln(u)$ with respect to $u$ is $\frac{1}{u}$.
By the chain - rule, the derivative of $\ln(1 + e^{12\theta})$ with respect to $\theta$ is $\frac{12e^{12\theta}}{1 + e^{12\theta}}$.

Step4: Calculate $\frac{dy}{d\theta}$

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

Answer:

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