find the derivative of y with respect to θ. y = ln(sec(ln(5θ))) \n$\frac{dy}{d\theta}=square$

Explanation:

Step1: Apply chain - rule for outer function

The outer function is $y = \ln(u)$ where $u=\sec(\ln(5\theta))$. The derivative of $\ln(u)$ with respect to $u$ is $\frac{1}{u}$. So, $\frac{dy}{du}=\frac{1}{\sec(\ln(5\theta))}$.

Step2: Differentiate the middle function

The middle function is $u = \sec(v)$ where $v=\ln(5\theta)$. The derivative of $\sec(v)$ with respect to $v$ is $\sec(v)\tan(v)$. So, $\frac{du}{dv}=\sec(\ln(5\theta))\tan(\ln(5\theta))$.

Step3: Differentiate the inner function

The inner function is $v=\ln(5\theta)$. Using the chain - rule for $\ln(5\theta)$ (or the fact that $\frac{d}{d\theta}\ln(a\theta)=\frac{1}{\theta}$ for $a > 0$), $\frac{dv}{d\theta}=\frac{5}{5\theta}=\frac{1}{\theta}$.

Step4: Use the chain - rule formula $\frac{dy}{d\theta}=\frac{dy}{du}\cdot\frac{du}{dv}\cdot\frac{dv}{d\theta}$

Substitute the values we found: $\frac{dy}{d\theta}=\frac{1}{\sec(\ln(5\theta))}\cdot\sec(\ln(5\theta))\tan(\ln(5\theta))\cdot\frac{1}{\theta}$. Simplify the expression: $\frac{dy}{d\theta}=\frac{\tan(\ln(5\theta))}{\theta}$.

Answer:

$\frac{\tan(\ln(5\theta))}{\theta}$