In order to obtain the second derivative, we will have to differentiate each component twice. We know how to differentiate the following:

$\displaystyle (x^n)''=n(n-1)x^{n-2}$.

We also have :

$\displaystyle (arctan(x))'=\frac{1}{1+x^2}$

and this gives:

$\displaystyle \{arctan(x)\}''=\frac{-2x}{(1+x^2)^2}$

For the constant component , we know that it is derivative is zero.

This gives us the solution that we are looking for:

$\displaystyle \vec{u}''=\left(n(n-1)x^{n-2},\frac{-2x}{(1+x^2)^2},0\right)$

By definition, to find the gradient vector , we will have to find the gradient components. We know that the gradient components are that partial derivatives.

We know that in our case we have :

$\displaystyle \frac{\partial f}{\partial x_{i}}=\frac{x_{i}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}$

To see this, fix all other variables and assume that you have only $\displaystyle x_{i}$ as the only variable.

Now we apply the given defintion , i.e,

$\displaystyle \nabla f =\left[\frac{\partial f}{\partial x_{1}} \cdots \frac{\partial f}{\partial x_{n}}\right]$

with :

$\displaystyle \frac{\partial f}{\partial x_{1}}=\frac{x_{1}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}$

$\displaystyle \frac{\partial f}{\partial x_{2}}=\frac{x_{2}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}$

$\displaystyle \frac{\partial f}{\partial x_{3}}=\frac{x_{3}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}$

$\displaystyle \vdots$

$\displaystyle \frac{\partial f}{\partial x_{n}}=\frac{x_{n}}{\sqrt{{x_{1}^2}+x_{2}^2+\cdots+x_{n}^2}}$

this gives us the solution .

We note first that :

$\displaystyle \frac{\partial f}{\partial x_{1}}=cos(x_{1}+2x_{2}+\cdots nx_{n})$

Using the Chain Rule where $\displaystyle x_{1}$ is the only variable here.

$\displaystyle \frac{\partial f}{\partial x_{2}}=2 cos(x_{1}+2x_{2}+\cdots nx_{n})$

Using the Chain Rule where $\displaystyle x_{2}$ is the only variable here.

Continuing in this fashion we have:

$\displaystyle \frac{\partial f}{\partial x_{n}}=ncos(x_{1}+2x_{2}+\cdots nx_{n})$

Again using the Chain Rule and assuming that $\displaystyle x_{n}$ is the variable and all the others are constant.

Now applying the given definition of the gradient we have the required result.

 $\displaystyle [cos(x_{1}+2x_{2}+\cdots nx_{n}) \quad 2cos(x_{1}+2x_{2}+\cdots nx_{n})\quad \cdots\quad ncos(x_{1}+2x_{2}+\cdots nx_{n})\]$

To find the derivative of this vector, all we need to do is to differentiate each component with respect to t.

Use the Power Rule and the Chain Rule when differentiating.

$\displaystyle (\sqrt{1+t^2})dt=(1+t^2)^\frac{1}{2}dt=\frac{1}{2}(1+t^2)^{-\frac{1}{2}}\cdot 2t=\frac{t}{\sqrt{t^2+1}}$  is the derivative of the first component.

$\displaystyle (2\sqrt{1+t^2})dt=2(1+t^2)^\frac{1}{2}dt=2\frac{1}{2}(1+t^2)^{-\frac{1}{2}}\cdot 2t=\frac{2t}{\sqrt{t^2+1}}$ of the second component.

$\displaystyle \vdots$

$\displaystyle (n\sqrt{1+t^2})dt=n(1+t^2)^\frac{1}{2}dt=n\frac{1}{2}(1+t^2)^{-\frac{1}{2}}\cdot 2t=\frac{nt}{\sqrt{t^2+1}}$ is the derivative of the last component . we obtain then:

$\displaystyle \left(\frac{t}{\sqrt{t^2+1}},\frac{2t}{\sqrt{t^2+1}},\cdots \frac{nt}{\sqrt{t^2+1}}\right)$

To find the derivative of u, we will have to find the derivative of every component.

Using the Chain Rule, we note that :

$\displaystyle aratan(t)'=\frac{1}{1+t^2}$

$\displaystyle artan(2t)'=\frac{2}{1+(2t)^2}=\frac{2}{1+4t^2}$

$\displaystyle artan(3t)'=\frac{3}{1+(3t)^2}=\frac{3}{1+9t^2}$

Continuing in the same fashion we have :

$\displaystyle artan((n-1)t)'=\frac{(n-1)}{1+((n-1)t)^2}=\frac{n-1}{1+(n-1)^2t^2}$

Ordering these components  pairwise , we have then:

$\displaystyle \vec{u}'=\left(\frac{1}{1+t^2}, \frac{2}{1+4t^2},\frac{3}{1+3t^2}, \cdots\quad ,\frac{n-1}{1+(n-1)^2t^2}\right)$

this is what we needed to show.

To obtain the desired result, we will have to differentiate the expression of u with respect to each variable and add the derivatives.

We have

$\displaystyle \frac{\partial u}{\partial x_{1}}=b_{1}^2e^{b_{1}^2x_{1}+b_{2}^2x_{2}+\cdots +b_{n}^2x_{n}}$

$\displaystyle \frac{\partial u}{\partial x_{2}}=b_{2}^2e^{b_{1}^2x_{1}+b_{2}^2x_{2}+\cdots +b_{n}^2x_{n}}$

$\displaystyle \frac{\partial u}{\partial x_{3}}=b_{3}^2e^{b_{1}^2x_{1}+b_{2}^2x_{2}+\cdots +b_{n}^2x_{n}}$

$\displaystyle \vdots$

$\displaystyle \frac{\partial u}{\partial x_{n}}=b_{n}^2e^{b_{1}^2x_{1}+b_{2}^2x_{2}+\cdots +b_{n}^2x_{n}}$

we can also write the above expression as :

$\displaystyle \frac{\partial u}{\partial x_{1}}=b_{1}^2u$

$\displaystyle \frac{\partial u}{\partial x_{2}}=b_{2}^2u$

$\displaystyle \vdots$

$\displaystyle \frac{\partial u}{\partial x_{n}}=b_{n}^2u$

Now summing up we have:

$\displaystyle \frac{\partial u}{\partial x_{1}}+\frac{\partial u}{\partial x_{2}}+\cdots +\frac{\partial u}{\partial x_{n}}=(b_{1}^2+b_{2}^2+\cdots b_{n}^2)u}}$

But we are given that :

$\displaystyle (b_{1}^2+b_{2}^2+\cdots b_{n}^2)=1$

This gives:

$\displaystyle \frac{\partial u}{\partial x_{1}}+\frac{\partial u}{\partial x_{2}}+\cdots +\frac{\partial u}{\partial x_{n}}=u$

One way to obtain the solution is that we first add the two functions to obtain $\displaystyle \vec{w}$ and we differentiate with each component to get the result.

 Adding the two functions we obtain:

$\displaystyle \vec{w}=(t+tcos(t), tsin(t)+t)$

Now we will differentialte each component to get the derivative of $\displaystyle \vec{w}$.

We have :

$\displaystyle \{t+tcos(t)\}'=1-tsin(t)+cos(t)$

$\displaystyle \{t+tsin(t)\}'=1+tcos(t)+sin(t)$

$\displaystyle \vec{w}'=((1-tsin(t)+cos(t),1+tcos(t)+sin(t))$

To obtain the derivative we will need to compute the derivative of each component. Note that we have a polynomial in this case.

We are also given that $\displaystyle n>m$ and $\displaystyle n> s$ with m,n and s positive integers.

We know that if given $\displaystyle f=t^n$, then $\displaystyle f^{[k]}=n(n-1)...(n-(k-1))t^{n-k}$

so for k=n, we have $\displaystyle [t^{n}]^{[n]}=n!$

and

$\displaystyle {\t^{m}\}^{[k]}=0$$\displaystyle t^{m}^{[n]}=0$ for m< n  .

In the same manner we have :

$\displaystyle t^{s}^{[n]}=0$ 

This shows that the derivative of order n is given by:

$\displaystyle (n!,0,0)$

We first have to determine the expression of the velocity and see what happens as we move toward infinity.

Let us first compute the derivative. To do that, we do it componentwize.

We have,

$\displaystyle (-e^{-t})'=e^{-t}$

$\displaystyle (artant)'=\frac{1}{1+t^2}$

$\displaystyle (t-\frac{1}{t})'=1+\frac{1}{t^2}$

Therefore the expression of the velocity .

Since we have $\displaystyle e^{-t}>0$,$\displaystyle \frac{1}{1+t^2}>0$ ,$\displaystyle 1+\frac{1}{t^2}>0$ for all t.

The velocity can never be zero.

To have $\displaystyle \vec{u}$ defined, we need to have all the components defined on the same interval, $\displaystyle ln(t)$ for $\displaystyle t>0$

$\displaystyle arctan(t)$ is defined for all t.

$\displaystyle \frac{1}{1-t}$ is defined if $\displaystyle t\neq 1$

This gives $\displaystyle (t>0)\cap(t\neq 1)\cap (-\infty, \infty )=(0,1)\cup (1,\infty)$