2. let $v(x)=xsin(x)$. find $v(x),v(x)$ and $v^{(2)}(x)$. 3. using the quotient rule, show that…

Explanation:

Step1: Find the first - derivative of $v(x)=x\sin(x)$

Use the product rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u = x$ and $v=\sin(x)$. The derivative of $u=x$ is $u^\prime = 1$, and the derivative of $v = \sin(x)$ is $v^\prime=\cos(x)$. So $v^\prime(x)=\sin(x)+x\cos(x)$.

Step2: Find the second - derivative of $v(x)$

Differentiate $v^\prime(x)=\sin(x)+x\cos(x)$ using the sum rule $(u + v)^\prime=u^\prime+v^\prime$. The derivative of $\sin(x)$ is $\cos(x)$, and for $x\cos(x)$ use the product rule again. Let $u = x$ and $v=\cos(x)$, then $u^\prime = 1$ and $v^\prime=-\sin(x)$. So the derivative of $x\cos(x)$ is $\cos(x)-x\sin(x)$. Then $v^{\prime\prime}(x)=\cos(x)+\cos(x)-x\sin(x)=2\cos(x)-x\sin(x)$.

Step3: Recall the quotient rule and the definitions of $\cot(\theta)$

The quotient rule is $\left(\frac{u}{v}\right)^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Since $\cot(\theta)=\frac{\cos(\theta)}{\sin(\theta)}$, let $u = \cos(\theta)$ and $v=\sin(\theta)$. Then $u^\prime=-\sin(\theta)$ and $v^\prime=\cos(\theta)$.

Step4: Apply the quotient rule to $\cot(\theta)$

$\frac{d}{d\theta}\cot(\theta)=\frac{-\sin(\theta)\sin(\theta)-\cos(\theta)\cos(\theta)}{\sin^{2}(\theta)}=\frac{-\left(\sin^{2}(\theta)+\cos^{2}(\theta)\right)}{\sin^{2}(\theta)}$.

Step5: Use the Pythagorean identity

Since $\sin^{2}(\theta)+\cos^{2}(\theta) = 1$, we have $\frac{d}{d\theta}\cot(\theta)=-\frac{1}{\sin^{2}(\theta)}=-\csc^{2}(\theta)$.

Answer:

$v^\prime(x)=\sin(x)+x\cos(x)$, $v^{\prime\prime}(x)=2\cos(x)-x\sin(x)$, and the proof that $\frac{d}{d\theta}\cot(\theta)=-\csc^{2}(\theta)$ is shown above.