# I Chain rule found in MIT video

1. Jul 14, 2016

### imsmooth

While solving an equation, the lecturer was using substitution in this video:

x=au was subbed in for Psi at timestamp 39:27
d/dx = (1/a)(d/du). I get that.
But then the second derivative is stated as being
d2/dx2 = (1/a2)(d2/du2)

How is it (1/a2) if we do not know if there is an "a" in the function "u"?

2. Jul 15, 2016

### Orodruin

Staff Emeritus
What do you mean? If you are fine with (d/dx) = (1/a)(d/du), how can you doubt (d/dx)^2 = (1/a^2)(d/du)^2?

3. Jul 15, 2016

### imsmooth

It is not the result squared. It is the second derivative.

4. Jul 15, 2016

### Orodruin

Staff Emeritus
Yes, which is exactly what I wrote. Taking the second derivative is equivalent to applying the derivative operator twice. This is standard operator notation.

5. Jul 15, 2016

### imsmooth

I guess I'm missing it or just thinking of it wrong:

x = au
dx = a du
dx2 = adu2 (since "a" is constant)

So, why is it a2?

6. Jul 15, 2016

### Orodruin

Staff Emeritus
This makes no sense whatsoever, it is not clear what you mean by $dx^2$. In standard notation, this would be either the differential of $x^2$ or the differential $dx$ squared. Neither behaves as you imply.

7. Jul 15, 2016

### imsmooth

My notation is off. What I was trying to ask is why is the second derivative of "x" not "a" times the second derivative of "u"? If "a" is not a variable of "u", where is the "a^2" coming from?

8. Jul 16, 2016

### Orodruin

Staff Emeritus
The second derivative of $x$ or with respect to $x$? Your original question was the derivative with respect to $x$.

The following might help: You agree that the first derivative of a function wrt $x$ follows the rule
$$\frac{df}{dx} = \frac{1}{a} \frac{df}{da}$$
if $x = au$ and $a$ is a constant. Now, the second derivative is just applying the first derivative twice and so
$$\frac{d^2f}{dx^2} = \frac{d}{dx}\frac{df}{dx} = \frac{d}{dx} \frac{1}{a}\frac{df}{du} = \frac{1}{a} \frac{d}{dx}\frac{df}{du}.$$
Now, see $df/du$ just as any function $g$. Regardless of what the function $g$ is, we have agreed that the first derivative is given by $dg/dx = (1/a) dg/du$ and so
$$\frac{1}{a} \frac{d}{dx}\frac{df}{du} = \frac{1}{a} \frac{dg}{dx} = \frac{1}{a^2}\frac{dg}{du} = \frac{1}{a^2} \frac{d}{du}\frac{df}{du} = \frac{1}{a^2} \frac{d^2f}{du^2}.$$

Also consider the following example: Let $f(x) = x^2$, then if $x = au$ we find that $df/dx = 2x$ and $d^2f/dx^2 = 2$. We also find that $d^2f/du^2 = d(x^2)/du^2 = d(a^2u^2)/du^2 = a^2d(u^2)/du^2 = 2a^2$. Clearly, the relation $d^2f/dx^2 = (1/u^2) d^2f/du^2$ is satisfied in this case.

9. Jul 16, 2016

### PeroK

The way I would look at it is simply:

$\frac{d}{du} = a \frac{d}{dx}$

And:

$\frac{d^2}{du^2} = \frac{d}{du} \frac{d}{du} = (a \frac{d}{dx})(a \frac{d}{dx}) = a^2 \frac{d^2}{dx^2}$

In fact, you could replace the derivatives by any linear operators:

If $U = aX$ then $U^2 = a^2 X^2$

For any linear operators $U$ and $X$

10. Jul 16, 2016

### imsmooth

Thanks for all the help. My issue I guess is that I was looking at it as the derivative of a constant function instead of it being a linear operator.