could someone please explain this proof to me? $d(x^2)=2xdx+dxdx$

Question

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

Answer 1

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

$$d(f(x_1,x_2,\cdots, x_n)) = \sum_{i=1}^n \frac{\partial f}{\partial x_i} dx_i.$$

Answer 2

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.

Answer 3

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.

Answer 4

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:

1. the initial square
2. two new long and thin rectangles $x \times dx$ (one vertical to the right of the original square, one horizontal above it), and
3. 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?

Answer 5

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 $$\Delta(x^4)=(x+\Delta x)^4-x^4=\\4x^3.\Delta x+6x^.(\Delta x)^2+4.(\Delta x)^3+(\Delta x)^4$$