let (g(x)=xsin(x)). find (g(x)) and (g(x)).\n\nlet the function (f) be defined by (f(x)=\frac{\tan(x)-1}{sec(…

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