Question

1. 0 / 1 points differentiate. f(w) = \frac{1 + \sec(w)}{1 - \sec(w)} f(w) =

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = 1+\sec(w)$ and $v = 1 - \sec(w)$. First, find $u^\prime$ and $v^\prime$. The derivative of $\sec(w)$ is $\sec(w)\tan(w)$, so $u^\prime=\sec(w)\tan(w)$ and $v^\prime=-\sec(w)\tan(w)$.

Step2: Substitute into quotient - rule formula

\[

$$\begin{align*} f^\prime(w)&=\frac{\sec(w)\tan(w)(1 - \sec(w))-(1 + \sec(w))(-\sec(w)\tan(w))}{(1 - \sec(w))^{2}}\\ &=\frac{\sec(w)\tan(w)-\sec^{2}(w)\tan(w)+\sec(w)\tan(w)+\sec^{2}(w)\tan(w)}{(1 - \sec(w))^{2}} \end{align*}$$

\]

Step3: Simplify the numerator

The terms $-\sec^{2}(w)\tan(w)$ and $\sec^{2}(w)\tan(w)$ cancel out. So the numerator becomes $2\sec(w)\tan(w)$.

Answer:

$\frac{2\sec(w)\tan(w)}{(1 - \sec(w))^{2}}$