Question

(a) use the quotient rule to differentiate the function
f(x)=\frac{\tan(x)-1}{sec(x)}.
f(x)=

Explanation:

Step1: Recall the quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. For $f(x)=\frac{\tan(x)-1}{\sec(x)}$, let $u = \tan(x)-1$ and $v=\sec(x)$.

Step2: Find $u'$ and $v'$

We know that the derivative of $\tan(x)$ is $\sec^{2}(x)$ and the derivative of a constant (in this case, $- 1$) is $0$, so $u'=\sec^{2}(x)$. The derivative of $\sec(x)$ is $\sec(x)\tan(x)$, so $v'=\sec(x)\tan(x)$.

Step3: Apply the quotient - rule

$$
LATEXBLOCK0
$$

Since $\sec^{2}(x)-\tan^{2}(x) = 1$, we have:

$$
LATEXBLOCK1
$$

Answer:

$\cos(x)+\sin(x)$