could someone please explain this proof to me? $d(x^2)=2xdx+dxdx$ I know that $d(xy)=ydx+xdy$ and $d(x^2)=2xdx+dxdx$
I want to calculate $d(x^3)$ and $d(x^4)$??
can anyone explain that how can I calculate it? and what is the name of this rule?
Regards
 A: What you have originally stated--$d(x^2)=2xdx + dxdx$--is not correct. It should be that $d(x^2)=2xdx$.
Formally, though, if you want to calculate it it's quite simple. It's just defined by
$$d(f(x_1,x_2,\cdots, x_n)) = \sum_{i=1}^n \frac{\partial f}{\partial x_i} dx_i.$$
A: I don't know how you know 
$$
d(xy) = ydx + xdy
$$
but I think I can explain what you want to know.
The identity you quote comes from thinking about a rectangle with width $x$ and length $y$ that you make a little bit bigger by changing $x$ by a tiny $dx$ and $y$ by a tiny $dy$. Then the area of the new rectangle is
$$
(x+dx)(y+dy) = xy + xdy + ydx + dxdy .
$$
That means the change in area is
$$
d(xy) = xdy + ydx + dxdy .
$$
Since $dxdy$ is the product of two tiny numbers it's very tiny, so we can ignore it. That gives you what you say you know. If you think of a square, for which $x=y$ then you get the formula in your title before you ignore the $(dx)^2$.
To get the results you want, expand $(x+dx)^3$ and then $(x+dx)^4$ using the binomial theorem (or just plain high school algebra). See which terms in the change are really tiny.
Note: none of this is rigorous formal calculus, but it is how calculus began, and it's a good way to train your intuition.
A: Apply the rules you know recursively.  I am writing these out in an extremely pedantic way:
$d(x^4) = d(x x^3) = x^3dx + xdx^3 = x^3dx + xd(x x^2) = x^3dx + xx^2dx + x^2dx^2 = 2x^3dx + x^3dx + x^3dx = 4x^3dx$
In general,
$dx^n = nx^{n - 1}dx$ which you can show by mathematical induction.  We already know the rule holds for $n = 1$, so suppose it holds for $n = m$, we will show that it holds also for $n = m + 1$:
$dx^{m + 1} = dxx^m = xdx^m + x^mdx = mx^mdx + x^mdx = (m + 1)x^mdx$
And hence it must hold for any finite $n$.  That's it.
A: A visual intuition
Remember that with $d(f(x))$ we mean how much $f(x)$ increases, if $x$ grows by a tiny amount $dx$. So, with $d(x^2)$ we mean how much the area of a square of side $x$ increases, if $x$ grows by a tiny amount $dx$.
Picture a square of side $x$. Then "extend" its sides upwards and rightwards by $dx$ to $x+dx$. Now you have a slightly larger square, formed by:


*

*the initial square

*two new long and thin rectangles $x \times dx$ (one vertical to the right of the original square, one horizontal above it), and

*a new tiny little $dx \times dx$ square in the upright corner. 


Those two rectangles and the little square are how much your $x^2$ increased as $x$ grew by the tiny amount $dx$:
$d(x^2)=2(x~dx)+(dx~dx)$.
What happens if you take a cube instead of a square?
A: Note that :to find $d(xy)$ 
$$\Delta(xy)=(x+\Delta x)(y+\Delta y)-xy=x\Delta y+y\Delta x+\Delta x\Delta y+xy-xy=\\x\Delta y+y\Delta x+\Delta x\Delta y \\$$so when  $\Delta y,\Delta x \to 0$ we have 
$$\large d(xy)=xdy+ydx+dxdy\\ x=y\\ \to d(xx)=d(x^2)=xdx+xdx+dx.dx=2xdx+(dx)^2 $$
now for the case of 1st method $$d(x^3)$$
$$=d(xx^2)=xd(x^2)+x^2dx+dx.d(x^2)=\\x(2x.dx+(dx)^2)+x^2.dx+dx.((2x.dx+(dx)^2)=\\2x^2.dx+x.(dx)^2+x^2.dx+2x(dx)^2+(dx)^3=\\3x^2.dx+3x(dx)^2+(dx)^3$$
second method
$$\Delta(x^3)=(x+\Delta x)^3-x^3=\\x^3.\Delta x+3x^2.\Delta x+(\Delta x)^3\\\Delta x \to 0 \\d(x^3)=3x^2.dx+3x(dx)^2+(dx)^3$$
then you can do it for $d(x^4)$ with two idea 
$$d(x^4)=d(x^2.x^2) \space or \space d(x.x^3)$$ or this way
$$\Delta(x^4)=(x+\Delta x)^4-x^4=\\4x^3.\Delta x+6x^.(\Delta x)^2+4.(\Delta x)^3+(\Delta x)^4$$
