Question

form 2 - topics 4 - 5: problem 7 (1 point) find f(x) for f(x)=\frac{4sec(x)}{15(1 + 2\tan(x))} f(x)=

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = \frac{4}{15}\sec(x)$ and $v=1 + 2\tan(x)$.

Step2: Find $u'$

The derivative of $\sec(x)$ is $\sec(x)\tan(x)$. So, $u'=\frac{4}{15}\sec(x)\tan(x)$.

Step3: Find $v'$

The derivative of $\tan(x)$ is $\sec^{2}(x)$. So, $v'=2\sec^{2}(x)$.

Step4: Apply quotient - rule

\[

$$\begin{align*} f'(x)&=\frac{\frac{4}{15}\sec(x)\tan(x)(1 + 2\tan(x))-\frac{4}{15}\sec(x)\times2\sec^{2}(x)}{(1 + 2\tan(x))^{2}}\\ &=\frac{\frac{4}{15}\sec(x)\tan(x)+\frac{8}{15}\sec(x)\tan^{2}(x)-\frac{8}{15}\sec^{3}(x)}{(1 + 2\tan(x))^{2}}\\ &=\frac{4\sec(x)\tan(x)+8\sec(x)\tan^{2}(x)-8\sec^{3}(x)}{15(1 + 2\tan(x))^{2}} \end{align*}$$

\]

Answer:

$\frac{4\sec(x)\tan(x)+8\sec(x)\tan^{2}(x)-8\sec^{3}(x)}{15(1 + 2\tan(x))^{2}}$