Show that $x\delta^{\prime}(x)=-\delta(x)$ The reasoning given in my book is as follows:

Multiply $x\delta^{\prime}(x)$ by $\phi(x)$ and integrate; where $\phi(x)$ is an arbitrary test function that is zero for large $\mid x \mid$ and $\delta^{\prime}(x)$ is the derivative of $\delta(x)$:
  $$\begin{align}\underbrace{\int_{x=-\infty}^{\infty}x\delta^{\prime}(x)\phi(x)\,\mathrm{d}x}_{\color{red}{\Large 1}}\\ & =\underbrace{-\Big(x\phi(x)\Big)^{\prime}\Big \lvert_{x=0}^{\infty}}_{\color{blue}{\Large 2}}\\ &
=\underbrace{-\Big(x\phi^{\prime}(x)+\phi(x)\Big)\Big \lvert_{x=0}^{\infty}}_{\color{#180}{\Large 3}}
\\&=\underbrace{-\phi(0)}_{\color{#F80}{\Large 4}}
\\&=\underbrace{-\int_{x=-\infty}^{\infty}\delta(x)\phi(x)\mathrm{d}x}_{\color{purple}{\Large 5}}\end{align}$$

So in short
$$\int_{x=-\infty}^{\infty}x\delta^{\prime}(x)\phi(x)\,\mathrm{d}x=-\int_{x=-\infty}^{\infty}\delta(x)\phi(x)\mathrm{d}x$$ 
Differentiation of both sides with respect to $x$ and then dividing both sides by $\phi(x)$ and the result follows: $$x\delta^{\prime}(x)=-\delta(x)$$  

I can't understand what steps were carried out to get from $\large\color{red}{1}$ to $\large\color{blue}{2}$. I tried by integrating by parts but this led to a dead end. 
I think that the product rule was used to get from $\large\color{blue}{2}$ to $\large\color{#180}{3}$.
By my calculation, to get from $\large\color{#180}{3}$ to $\large\color{#F80}{4}$ there should not be a negative sign in front of $\phi(0)$. Since $$-\Big(x\phi^{\prime}(x)+\phi(x)\Big)\Big\lvert_{x=0}^{\infty}=-\Big((\infty\cdot 0+0)-\big(0\cdot \phi^{\prime}(0)+\phi(0)\big)\Big)=-\big(-\phi(0)\big)=\phi(0)$$
I have absolutely no idea how to get from $\large\color{#F80}{4}$ to $\large\color{purple}{5}$. But I guess this depends on whether the previous step is correct.

Could someone please give me some hints or tips or better yet show the intermediate steps that were omitted from the book?
Thank you.

EDIT:
Please ignore all of the above as it is wrong due to a book error. A special thanks goes to @mjqxxxx , @Augustin and @ForgotALot for pointing this out to me.
I apologize for the inconvenience.   
The (correct) reasoning given is as follows:

Multiply $x\delta^{\prime}(x)$ by $\phi(x)$ and integrate; where $\phi(x)$ is an arbitrary test function that is zero for large $\mid x \mid$ and $\delta^{\prime}(x)$ is the derivative of $\delta(x)$:
  $$\begin{align}\underbrace{\int_{x=-\infty}^{\infty}x\delta^{\prime}(x)\phi(x)\,\mathrm{d}x}_{\color{red}{\Large 1}}\\ & =\underbrace{-\Big(x\phi(x)\Big)^{\prime}\Big \lvert_{x=0}}_{\color{blue}{\Large 2}}\\ &
=\underbrace{-\Big(x\phi^{\prime}(x)+\phi(x)\Big)\Big \lvert_{x=0}}_{\color{#180}{\Large 3}}
\\&=\underbrace{-\phi(0)}_{\color{#F80}{\Large 4}}
\\&=\underbrace{-\int_{x=-\infty}^{\infty}\delta(x)\phi(x)\mathrm{d}x}_{\color{purple}{\Large 5}}\end{align}$$

So in short
$$\int_{x=-\infty}^{\infty}x\delta^{\prime}(x)\phi(x)\,\mathrm{d}x=-\int_{x=-\infty}^{\infty}\delta(x)\phi(x)\mathrm{d}x$$ 
Differentiation of both sides with respect to $x$ and then dividing both sides by $\phi(x)$ and the result follows: $$x\delta^{\prime}(x)=-\delta(x)$$
Could someone please give me some hints or tips or better yet show the intermediate steps that were omitted from the book?
 A: You don't say what your book is. Let me first give some background and then address your book's derivation.  For background I look at Rudin's Functional Analysis ch. 6 which discusses distributions, which I believe to be the usual framework in which mathematicians view "the Dirac delta function."  There's also Yosida's Functional Analysis, Hörmander volume 1, the old Schwartz book in French, etc.  
Background.  In this framework, the "delta function" is actually a continuous linear function from $\mathscr{D}$ to $\mathbb{C}$, where  $\mathscr{D}$ is the $C^\infty$ functions on $\mathbb{R}$ with compact support, endowed with a suitable topology.  By definition (Rudin 6.9), $\delta(\phi) = \phi(0)$ for any $\phi\in\mathscr{D}$.  Also by definition, the first derivative $\delta'$ is defined by $\delta'(\phi)=-\delta(\phi')=-\phi'(0)$ (Rudin 6.12) for any $\phi\in\mathscr{D}$.  The value $\phi'(0)$ can alternately be written $d\phi(x)/dx|_{x=0}$. 
The product of a function mapping $x$ to $x$ by a distribution is, yet again by definition (Rudin 6.15), a distribution satisfying $(x\delta)(\phi)=\delta(x\phi)$ for any $\phi\in\mathscr{D}$, where we use $x$ to denote the function which maps $x$ to $x$ (others would get fancier and say $Id$ for identity or something like that).
Now, putting these definitions together, $(x\delta')(\phi)=\delta'(x\phi)=-\delta(d(x\phi(x))/dx)$.  Now, $x\phi$ is a good ol' $C^\infty$ function, the product rule applies, and so $d(x\phi(x))/dx= \phi(x)+x\phi'(x)$.  Substituting in the prior formula and remembering that $\delta(\phi_1)=\phi_1(0)$ for all $\phi_1\in\mathscr{D}$, we get $(x\delta')(\phi)=-\phi(0)=-\delta(\phi)$.  Since $\phi\in\mathscr{D}$ was arbitrary, this means $x\delta'=-\delta$.
If $\Lambda$ is a distribution and $\phi\in\mathscr{D}$, one sometimes writes 
$$\Lambda(\phi)=\int_{-\infty}^\infty \Lambda(x)\phi(x)\,dx$$
That is what your book seems to be doing.  However, it helps to keep the definitions and the distinction between distributions and functions clearly in mind, especially at the beginning when one is learning about them.
The answer to your question.  Turning now at last to your book's reasoning, the big oddity is the $|_{x=0}^\infty$ in step 2.  Remember that step 1 is just the intuitively helpful (or not) way of writing the action of a distribution on a test function as an integral.  Suppose it were a real integral of good ol' functions on $\mathbb{R}$:  then integration by parts would apply to it, and then one would apply something like $$\int_{-\infty}^\infty f(x)g'(x)\,dx=f(x)g(x)|^\infty_{-\infty}-\int_{-\infty}^\infty f'(x)g(x)\,dx.$$  Now, here $f(x)=x\phi(x)$ and $g(x)=\delta(x)$.  Because $\phi(x)$ is a test function $\in \mathscr{D}$, it equals to $0$ outside a compact set, and so the product $f(x)g(x)|_{-\infty}^\infty$ should equal $0$ with the identifications of $f$ and $g$ we've made (plus, as conventionally understood, $g(x)=\delta(x)=0$ for $x\ne 0$).  Then, to eliminate the second integral sign, one would apply the definition of $\delta$, but the book does it wrong by using $|_{x=0}^\infty$ rather than the correct $|_{x=0}$.  The book's mistake introduces a sign error since it's subtracting the value at $x=0$ rather than just plunking it into the formula.  Step 4, as you correctly point out, makes a further sign error, and the two cancel out.
In sum, your book's derivation is wrong at step 2, and it corrects the error at step 4 by introducing a further error.
