Why isn't the notation for the second derivative $\frac{{d y}^{2}}{d x}$ $dy^2/dx$ ?

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Why isn't the notation for the second derivative $\frac{{d y}^{2}}{d x}$ $dy^2/dx$ ?

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Answer 1 · 2017-11-18T19:45:44

See below.

Explanation:

Probably because $\frac{{d y}^{2}}{d x}$ $(dy^2)/dx$ would be read as the derivative of $y^{2}$ $y^2$ in respect of $x$ $x$ .

The general notation $\frac{d^{2} y}{{d x}^{2}}$ $(d^2y)/dx^2$ could be misconstrued as the derivative in respect of $x^{2}$ $x^2$ , but then you can find lots of flaws in mathematical notations, but we just have to accept that this is how they have been defined.

Answer 2 · 2017-11-18T20:11:19

Please see below.

Explanation:

For $f (x)$ $f(x)$ a function of $x$ $x$ , we can denote the derivative in several ways.

On family of notations uses a "differential operator".
This is a notation that tells us "take the derivative".

For example $D_{x} (5 x^{2} + 3)$ $D_x(5x^2+3)$ means "The derivative, with respect to $x$ $x$ , of the function $f (x) = 5 x^{2} + 3$ $f(x)=5x^2+3$ . SO $D_{x} (5 x^{2} + 3) = 10 x$ $D_x(5x^2+3) = 10x$ .

$\frac{d}{d x}$ $d/dx$ is another notation for "the derivative with respect to $x$ $x$ of . . . #

So $\frac{d}{d x} (5 x^{2} + 3)$ $d/dx(5x^2+3)$ means the same thing as $D_{x} (5 x^{2} + 3)$ $D_x(5x^2+3)$ .

To take the second derivative, we take the derivative twice:

$D_{x} (D_{x} (5 x^{2} + 3))$ $D_x(D_x(5x^2+3))$ is written $D_{x}^{2} (5 x^{2} + 3)$ $D_x^2(5x^2+3)$

$\frac{d}{d x} (\frac{d}{d x} (5 x^{2} + 3))$ $d/dx(d/dx(5x^2+3))$ could, perhaps, be written ${(\frac{d}{d x})}^{2} (5 x^{2} + 3)$ $(d/dx)^2(5x^2+3)$ , but the standard notation is $\frac{d^{2}}{{d x}^{2}} (5 x^{2} + 3)$ $d^2/dx^2(5x^2+3)$ .

If we let $y = 5 x^{2} + 3$ $y = 5x^2+3$ , then these examples look like:

First derivative: $D_{x} (y)$ $D_x(y)$ and $\frac{d}{d x} (y)$ $d/dx(y)$ which is also written $\frac{d y}{d x}$ $dy/dx$ .

Second derivative: $D_{x}^{2} (y)$ $D_x^2(y)$ and $\frac{d^{2}}{{d x}^{2}} (y)$ $d^2/dx^2(y)$ which is also written $\frac{d^{2} y}{{d x}^{2}}$ $(d^2y)/dx^2$ .

In this notation, we do not think of $d x$ $dx$ as $d$ $d$ times $x$ $x$

Instead, we are thinking of $d x$ $dx$ as a single quantity. That is why we do NOT write $\frac{d^{2}}{{(d x)}^{2}} (y)$ $d^2/(dx)^2(y)$

Additional Note

I would interpret $\frac{{d y}^{2}}{d x}$ $dy^2/dx$ as $\frac{d}{d x} (y^{2})$ $d/dx(y^2)$ . Using the chain rule, I would evaluate to get $\frac{d}{d x} (y^{2}) = 2 y \cdot \frac{d y}{d x}$ $d/dx(y^2) = 2y * dy/dx$ .

Using $y = 5 x^{2} + 3$ $y=5x^2+3$ again, we have $\frac{d}{d x} {(5 x^{2} + 3)}^{2} = 2 (5 x^{2} + 3) \cdot 5 = 10 (5 x^{2} + 3)$ $d/dx(5x^2+3)^2 = 2(5x^2+3)*5 = 10(5x^2+3)$

Note 2

${(\frac{d y}{d x})}^{2}$ $(dy/dx)^2$ means the square of the derivative of $y$ $y$ .
Using $y = 5 x^{2} + 3$ $y=5x^2+3$ again, we have

${(\frac{d y}{d x})}^{2} = {(10 x)}^{2} = 100 x^{2}$ $(dy/dx)^2 = (10x)^2 = 100x^2$ .

This kind of expression can come up if using implicit differentiation to find 2nd or higher derivatives. (If we don't pause to solve for $\frac{d y}{d x}$ $dy/dx$ before our second differentiation.)

Answer 3 · 2017-11-18T21:13:20

See below.

Explanation:

$\frac{d}{d x} (\frac{d y}{d x}) = \frac{d^{2} y}{{d x}^{2}}$ $d/(dx)(dy/dx) = (d^2y)/(dx^2)$
$\frac{d}{d x} (\frac{d}{d x}) y = \frac{d^{2}}{{d x}^{2}} y = \frac{d^{2} y}{{d x}^{2}}$ $d/(dx)(d/dx)y = (d^2)/(dx^2)y = (d^2y)/(dx^2)$

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