# Double derivative definition question

1. Mar 4, 2009

### transgalactic

a derivative of function is marked as
$$\frac{\mathrm{df} }{\mathrm{d} x}$$
why the double derivative is marked as
$$\frac{\mathrm{d^2f} }{\mathrm{d} x^2}$$

i tried to build it using the original definition
$$\frac{d\frac{\mathrm{df} }{\mathrm{dx}} }{dx}$$
its not working
??

2. Mar 4, 2009

### phreak

It's a silly question, and is more due to notation than to a derivation. If you wanted to do a derivation, however, it would go along the lines of:

$$\frac{d}{dx} \frac{d}{dx} f = \left(\frac{d}{dx}\right)^2 f = \frac{d^2}{(dx)^2}f$$ and we omit the parentheses from the denominator.

3. Mar 4, 2009

### HallsofIvy

Staff Emeritus
You are only asking about the notation? It's just notation. I don't know what you mean by trying to "build it". (And in English it's "second" derivative, not "double" derivative.)

4. Mar 4, 2009

### transgalactic

ok you got
$$\frac{d^2}{(dx)^2}f$$
its not
$$\frac{\mathrm{d^2f} }{\mathrm{d} x^2}$$

5. Mar 4, 2009

### phreak

No, they're the same thing. $$\frac{d}{dx}$$ and $$\frac{d^2}{(dx)^2}$$ are operators on functions. It doesn't matter if you put it in the numerator or off to the side, it means the same thing. As for dropping the parentheses, we have two ways we could possibly interpret $$dx^2$$. Either $$d(x^2)$$ or $$(dx)^2$$. But $$x^2$$ is completely irrelevant to second differentiation, so it has to be the latter.