Question

1. if ( f(x)=sin(cos(\tan x)) ), find ( f(x) )
2. differentiate ( y = e^{sec 3\theta} )

Explanation:

Step1: Let \(u = \sec(3\theta)\)

So \(y = e^{u}\). By the chain rule \(\frac{dy}{d\theta}=\frac{dy}{du}\cdot\frac{du}{d\theta}\).
First, find \(\frac{dy}{du}\):
Since \(y = e^{u}\), then \(\frac{dy}{du}=e^{u}\).

Step2: Find \(\frac{du}{d\theta}\)

Let \(v = 3\theta\), so \(u=\sec(v)\).
By the chain rule \(\frac{du}{d\theta}=\frac{du}{dv}\cdot\frac{dv}{d\theta}\).
Since \(u = \sec(v)\), \(\frac{du}{dv}=\sec(v)\tan(v)\). And since \(v = 3\theta\), \(\frac{dv}{d\theta}=3\).
So \(\frac{du}{d\theta}=3\sec(3\theta)\tan(3\theta)\).

Step3: Combine the results

\(\frac{dy}{d\theta}=\frac{dy}{du}\cdot\frac{du}{d\theta}\).
Substitute \(\frac{dy}{du}=e^{u}\) and \(\frac{du}{d\theta}=3\sec(3\theta)\tan(3\theta)\) with \(u = \sec(3\theta)\).
We get \(\frac{dy}{d\theta}=e^{\sec(3\theta)}\cdot3\sec(3\theta)\tan(3\theta)\)

Answer:

\(y^\prime=3e^{\sec(3\theta)}\sec(3\theta)\tan(3\theta)\)