# Flip dx/dy according to chain rule

1. Jun 12, 2010

### Moogie

Hi

I've just been reading something which is essentially how to work out what the deriviative of y=b^x is.

Basically the explanation gets to the point which I understand and says

$$\frac{dx}{dy} = \frac{1}{yln(b)}$$

It then says because of the chain rule you can simply flip this to get

$$\frac{dy}{dx} = yln(b)$$

I have used the chain rule and thought I understood it but perhaps I don't because I don't see how the chain rule lets you do the 'flip' just described.

Kind regards

2. Jun 12, 2010

### tiny-tim

Hi Moogie!

(try using the X2 tag just above the Reply box )

If y is a function of x, and x is a function of t, then of course the chain rule gives

dy/dt = dy/dx dx/dt​

Now put y = t … then we get

1 = dy/dy = dy/dx dx/dy

3. Jun 12, 2010

### Moogie

Hi

I think you are trying to show me that dy/dx * dx/dy = 1 but I don't understand why (my limitation, not your explanation).

Could you try and explain it again from a different perspective? I didn't get the y=t thing. I can see how the 'symbols' make sense but I have no understanding or intuition of whats going on behind the symbols and thats more important to me.

thank you

4. Jun 12, 2010

### Staff: Mentor

If the product of two expressions is 1, the expressions are reciprocals of each other.

In tiny-tim's work, if y = t, then dy/dt = 1.

5. Jun 13, 2010

### Moogie

Hi

I understand that if the product of 2 expressions is 1 then they are reciprocals. I failed to understand where the chain rule came into it. How does the chain rule tell me that dy/dx * dx/dy = 1?

I was hoping someone might be able to explain it to me from a different perspective (unless it is that simple that there isn't another perspective)

6. Jun 13, 2010

### tiny-tim

Hi Moogie!

I don't think there is any different perspective

Do you accept that dy/dt = dy/dx dx/dt comes from the chain rule?

If so, the result does come from putting t = y.

7. Jun 13, 2010

### HallsofIvy

Staff Emeritus
Then, what, exactly, do you understand as the "chain rule"?

8. Jun 13, 2010

### Moogie

Hi Tiny Tim

I do see that dy/dt = dy/dx dx/dt from the chain rule. Perhaps I am missing what it is you are doing when you set t = y

thanks

9. Jun 13, 2010

### vela

Staff Emeritus
Say you have a function y=f(x) and it has an inverse x=g(y). Then dy/dx=f'(x) and dx/dy=g'(y). Because f and g are inverses, you have x=g(f(x)). If you differentiate this, you get, using the chain rule,

$$x=g(f(x)) \hspace{0.3in} \Rightarrow \hspace{0.3in} 1=g'(f(x))f'(x)=g'(y)f'(x)=\frac{dx}{dy}\frac{dy}{dx}$$

10. Jun 14, 2010

### Moogie

The explanation with inverses made more sense to me. But now I'm thinking perhaps I don't understand the chain rule after all.

When you go from x= g(f(x)) => 1 = g'(f(x))f'(x)

are you differentiating both sides wrt x? I'm not that familiar with the g/f version of the chain rule though I have seen it. Only in american books though. The UK doesn't seem to know about it!

I think maybe I'm getting confused because I'm so used to seeing y is a function f of x, and not x as a function of anything.

11. Jun 14, 2010

### vela

Staff Emeritus
Yes, both sides are differentiated with respect to x. It might be a little clearer to start with x=g(y), then when you differentiate, you get

$$1 = \frac{dg}{dy} \frac{dy}{dx}$$

which is the form of the chain rule you're familiar with. Then since g is just x, you get

$$1 = \frac{dx}{dy} \frac{dy}{dx}$$

12. Jun 14, 2010

### HallsofIvy

Staff Emeritus
Well, that's because the Brit's don't like to admit Leibniz had anything to do with Calculus and only use Newton's notation!

Newton's notation: f'(x).
Leibniz' notation: df/dx.

While the derivative is NOT a fraction, it is the limit of a fraction so it has "fraction properties"- Leibniz' notation allows you to think of it as a fraction. In particular, the chain rule becomes
$$\frac{df}{dx}= \frac{df}{dy}\frac{dy}{dx}$$
where y is any function of x- it looks like we are just cancelling the "dy"s.

Also if y= f(x) and, inverting, x= g(y), then
$$\frac{dx}{dy}= \frac{dg}{dy}= \frac{1}{\frac{df}{dx}}= \frac{1}{\frac{dy}{dx}}$$
just "inverting the fraction" (with the understanding, again, that the derivative isn't really a "fraction", it can just be treated like one!).