Question

let (g(x)=xsin(x)). find (g(x)) and (g(x)).

let the function (f) be defined by (f(x)=\frac{\tan(x)-1}{sec(x)}).
(a) use the quotient rule to differentiate the function (f(x)).
(b) simplify the expression for (f(x)) by writing it in terms of (sin(x)) and (cos(x)), and then find (f(x)).
(c) show that your answers to parts (a) and (b) are equivalent?
yes
no

Explanation:

Step1: Find $g'(x)$ using product - rule

The product - rule states that if $y = u\cdot v$, then $y'=u'v + uv'$. For $g(x)=x\sin(x)$, let $u = x$ and $v=\sin(x)$. Then $u' = 1$ and $v'=\cos(x)$. So $g'(x)=1\cdot\sin(x)+x\cdot\cos(x)=\sin(x)+x\cos(x)$.

Step2: Find $g''(x)$ by differentiating $g'(x)$

Differentiate $g'(x)=\sin(x)+x\cos(x)$ term - by - term. The derivative of $\sin(x)$ is $\cos(x)$, and for the second term $x\cos(x)$ (using product - rule with $u = x$, $u' = 1$, $v=\cos(x)$, $v'=-\sin(x)$), we get $1\cdot\cos(x)+x\cdot(-\sin(x))$. Then $g''(x)=\cos(x)+\cos(x)-x\sin(x)=2\cos(x)-x\sin(x)$.

Step3: Differentiate $f(x)=\frac{\tan(x)-1}{\sec(x)}$ using quotient - rule for part (a)

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u=\tan(x)-1$, $u'=\sec^{2}(x)$, $v = \sec(x)$, $v'=\sec(x)\tan(x)$. So $f'(x)=\frac{\sec^{2}(x)\cdot\sec(x)-(\tan(x) - 1)\cdot\sec(x)\tan(x)}{\sec^{2}(x)}=\frac{\sec^{3}(x)-\sec(x)\tan^{2}(x)+\sec(x)\tan(x)}{\sec^{2}(x)}=\sec(x)-\frac{\tan^{2}(x)}{\sec(x)}+\frac{\tan(x)}{\sec(x)}$.

Step4: Simplify $f(x)$ in terms of $\sin(x)$ and $\cos(x)$ for part (b)

Since $\tan(x)=\frac{\sin(x)}{\cos(x)}$ and $\sec(x)=\frac{1}{\cos(x)}$, $f(x)=\frac{\frac{\sin(x)}{\cos(x)}-1}{\frac{1}{\cos(x)}}=\sin(x)-\cos(x)$. Then $f'(x)=\cos(x)+\sin(x)$.

Step5: Show equivalence for part (c)

Rewrite $f'(x)$ from part (a) in terms of $\sin(x)$ and $\cos(x)$. $\sec(x)-\frac{\tan^{2}(x)}{\sec(x)}+\frac{\tan(x)}{\sec(x)}=\frac{1}{\cos(x)}-\frac{\frac{\sin^{2}(x)}{\cos^{2}(x)}}{\frac{1}{\cos(x)}}+\frac{\frac{\sin(x)}{\cos(x)}}{\frac{1}{\cos(x)}}=\frac{1}{\cos(x)}-\frac{\sin^{2}(x)}{\cos(x)}+\frac{\sin(x)}{\cos(x)}=\frac{1 - \sin^{2}(x)+\sin(x)}{\cos(x)}=\frac{\cos^{2}(x)+\sin(x)}{\cos(x)}$. This is wrong. Let's start over for part (a) simplification:
\[

$$\begin{align*} f'(x)&=\frac{\sec^{2}(x)\cdot\sec(x)-(\tan(x) - 1)\cdot\sec(x)\tan(x)}{\sec^{2}(x)}\\ &=\sec(x)-\tan(x)\frac{\tan(x) - 1}{\sec(x)}\\ &=\frac{1}{\cos(x)}-\frac{\sin(x)}{\cos(x)}\cdot\frac{\frac{\sin(x)}{\cos(x)}-1}{\frac{1}{\cos(x)}}\\ &=\frac{1}{\cos(x)}-\frac{\sin(x)}{\cos(x)}\cdot(\frac{\sin(x)}{\cos(x)}-1)\\ &=\frac{1}{\cos(x)}-\frac{\sin^{2}(x)}{\cos^{2}(x)}+\frac{\sin(x)}{\cos(x)}\\ &=\frac{\cos(x)-\sin^{2}(x)+\sin(x)\cos(x)}{\cos^{2}(x)}\\ \end{align*}$$

\]
Another way:
\[

$$\begin{align*} f'(x)&=\frac{\sec^{2}(x)\cdot\sec(x)-(\tan(x) - 1)\cdot\sec(x)\tan(x)}{\sec^{2}(x)}\\ &=\frac{\sec(x)(\sec^{2}(x)-\tan^{2}(x)+\tan(x))}{\sec^{2}(x)}\\ &=\frac{\sec(x)(1 + \tan(x))}{\sec^{2}(x)}\\ &=\frac{1+\frac{\sin(x)}{\cos(x)}}{\frac{1}{\cos(x)}}=\cos(x)+\sin(x) \end{align*}$$

\]

Answer:

$g'(x)=\sin(x)+x\cos(x)$
$g''(x)=2\cos(x)-x\sin(x)$
$f'(x)$ (from part (a)) $=\cos(x)+\sin(x)$
$f'(x)$ (from part (b)) $=\cos(x)+\sin(x)$
(c) Yes