Flip dx/dy according to chain rule

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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 let's you do the 'flip' just described.

Any comments greatly welcomed

Kind regards
 
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Hi Moogie! :smile:

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

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 :wink:
 
Hi

Thanks for your reply but I don't understand how that answers my question.

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 what's going on behind the symbols and that's more important to me.

thank you
 
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.
 
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)
 
Hi Moogie! :smile:

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.
 
Moogie said:
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)
Then, what, exactly, do you understand as the "chain rule"?
 
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
 
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
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
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
Moogie said:
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!
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! :biggrin::devil:

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!).

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.
 
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