Question

q5 (1 point)
if $f(x) = |\tan(e^x)|$, then $f(x) =$
$\tan(e^x)$
$e^x \tan(e^x)$
$\frac{(\sec(e^x))^2}{|\tan(e^x)|}e^x$
$\frac{\tan(e^x)}{|\tan(e^x)|}e^x$
$\frac{\tan(e^x)}{|\tan(e^x)|}e^x(\sec(e^x))^2$

Explanation:

Step1: Recall derivative of absolute value

For $g(x) = |h(x)|$, $g'(x) = \frac{h(x)}{|h(x)|}h'(x)$ (where $h(x)
eq0$). Here $h(x)=\tan(e^x)$.

Step2: Find derivative of $h(x)=\tan(e^x)$

Use chain rule: Let $u=e^x$, $\frac{d}{dx}\tan(u)=\sec^2(u)\cdot u'$.
$u' = \frac{d}{dx}e^x = e^x$, so $h'(x)=\sec^2(e^x)\cdot e^x$.

Step3: Combine using absolute value derivative

Substitute into $g'(x)$:
$f'(x) = \frac{\tan(e^x)}{|\tan(e^x)|}\cdot \sec^2(e^x)\cdot e^x$
Use identity $\sec^2(x)=1+\tan^2(x)$, and note $\tan(x)\sec^2(x)=\frac{\tan(x)\sec^2(x)|\tan(x)|}{|\tan(x)|}$ is unnecessary; instead, simplify using $\tan(x)\sec^2(x)=\frac{\tan(x)(1+\tan^2(x))}{|\tan(x)|}e^x$ is not needed. Instead, recognize $\frac{\tan(u)}{|\tan(u)|}\sec^2(u)e^x = \frac{\tan(u)\sec^2(u)}{|\tan(u)|}e^x$ is equivalent to $\frac{\tan(e^x)}{|\tan(e^x)|}e^x(\sec(e^x))^2$? No, correct simplification: $\frac{d}{dx}|tan(e^x)| = \frac{tan(e^x)}{|tan(e^x)|}\cdot \frac{d}{dx}tan(e^x) = \frac{tan(e^x)}{|tan(e^x)|}\cdot e^x sec^2(e^x)$ which is $\frac{\tan(e^x)}{|\tan(e^x)|}e^x(\sec(e^x))^2$. Wait, no, alternative: $\frac{\tan(u)}{|\tan(u)|} = \frac{|\tan(u)|}{\tan(u)}$? No, $\frac{d}{dx}|u| = \frac{u}{|u|}u'$. So $u=tan(e^x)$, $u'=e^x sec^2(e^x)$. So $f'(x)=\frac{tan(e^x)}{|tan(e^x)|}\cdot e^x sec^2(e^x) = e^x \frac{tan(e^x)sec^2(e^x)}{|tan(e^x)|}$. But also, $\sec^2(e^x)=\frac{1}{\cos^2(e^x)}$, and $\frac{tan(e^x)}{|tan(e^x)|}=\frac{sin(e^x)/cos(e^x)}{|sin(e^x)/cos(e^x)|}=\frac{sin(e^x)}{|sin(e^x)|}$ (since $cos(e^x)$ is positive or negative, but $|cos(e^x)|=|cos(e^x)|$, so $\frac{1/cos(e^x)}{1/|cos(e^x)|}=\frac{|cos(e^x)|}{cos(e^x)}$, so $\frac{tan(e^x)}{|tan(e^x)|}=\frac{sin(e^x)|cos(e^x)|}{|sin(e^x)|cos(e^x)}$). But better to use $\frac{tan(u)}{|tan(u)|}sec^2(u) = \frac{tan(u)sec^2(u)}{|tan(u)|} = \frac{sec^2(u)}{|tan(u)|}tan(u) = \frac{1/cos^2(u)}{|sin(u)/cos(u)|}\cdot \frac{sin(u)}{cos(u)} = \frac{1}{|sin(u)|cos(u)}\cdot \frac{sin(u)}{cos(u)} = \frac{sin(u)}{|sin(u)|cos^2(u)}$. No, correct step: $\frac{d}{dx}|tan(e^x)| = \frac{tan(e^x)}{|tan(e^x)|}\cdot e^x sec^2(e^x) = e^x \frac{tan(e^x)sec^2(e^x)}{|tan(e^x)|}$ which is the last option. Wait, no, wait: $\frac{d}{dx}|u| = \frac{u}{|u|}u'$. So $u=tan(e^x)$, $u'=e^x sec^2(e^x)$. So $f'(x)=\frac{tan(e^x)}{|tan(e^x)|}\cdot e^x sec^2(e^x) = e^x \frac{tan(e^x)}{|tan(e^x)|} sec^2(e^x)$ which is $\frac{\tan(e^x)}{|\tan(e^x)|}e^x(\sec(e^x))^2$.

Wait, another way: Let $y=|tan(e^x)|$, then $y^2=tan^2(e^x)$. Differentiate both sides: $2y y' = 2 tan(e^x) \cdot e^x sec^2(e^x)$. Then $y' = \frac{tan(e^x) e^x sec^2(e^x)}{y} = \frac{tan(e^x) e^x sec^2(e^x)}{|tan(e^x)|}$ which is $\frac{\tan(e^x)}{|\tan(e^x)|}e^x(\sec(e^x))^2$.

Answer:

$\frac{\tan(e^{x})}{|\tan(e^{x})|}e^{x}(\sec(e^{x}))^{2}$