homework5: problem 21 (1 point) find $f(x)$ for $f(x)=sec(x)\tan(x)$ $f(x)=$

Explanation:

Step1: Recall product - rule

The product - rule states that if $y = u(x)v(x)$, then $y'=u'(x)v(x)+u(x)v'(x)$. Here, $u(x)=\sec(x)$ and $v(x)=\tan(x)$.

Step2: Find derivatives of $u(x)$ and $v(x)$

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

Step3: Apply product - rule

$f'(x)=u'(x)v(x)+u(x)v'(x)=\sec(x)\tan(x)\cdot\tan(x)+\sec(x)\cdot\sec^{2}(x)$.

Step4: Simplify the expression

$f'(x)=\sec(x)\tan^{2}(x)+\sec^{3}(x)=\sec(x)(\tan^{2}(x)+\sec^{2}(x))$. Using the identity $\tan^{2}(x)+ 1=\sec^{2}(x)$, we can also write $f'(x)=2\sec^{3}(x)-\sec(x)$.

Answer:

$2\sec^{3}(x)-\sec(x)$