You are Here: Home >< Maths

# Inverse function derivative watch

1. So

I understand this can be shown from the identity map and just differentiating and making the subject.

But this looked familiar when my lecture notes said it's the same as if you let

Now I'm probably being stupid but, wouldn't that make the above ?
2. Can you explain further as I see it, if:

Differentiate with respect to y:

Therefore:

------------------------------------------------------------------------------------------------

Off topic: Is this to do with the "Inverse Function Theorem"?
3. (Original post by simon0)
Can you explain further as I see it, if:

Differentiate with respect to y:

Therefore:

------------------------------------------------------------------------------------------------

Off topic: Is this to do with the "Inverse Function Theorem"?
How did you get the last line From what's above?

It might be, but I'm just reading the rules for differentiation.
4. I believe you are missing the last part of the equation, the F and exponent can't just disappear and should be included, so you have 2/3's of the equation. Apologies if I am incorrect and on the wrong track, its been a while since I had quadratic equations, Calc 1 & 2, etc.

(Original post by NotNotBatman)
So

I understand this can be shown from the identity map and just differentiating and making the subject.

But this looked familiar when my lecture notes said it's the same as if you let

Now I'm probably being stupid but, wouldn't that make the above ?
5. (Original post by NotNotBatman)
How did you get the last line From what's above?

It might be, but I'm just reading the rules for differentiation.
I applied the reciprocal of dx/dy (so I did 1/(dx/dy) ). :-)
6. (Original post by luq_ali)
I believe you are missing the last part of the equation, the F and exponent can't just disappear and should be included, so you have 2/3's of the equation. Apologies if I am incorrect and on the wrong track, its been a while since I had quadratic equations, Calc 1 & 2, etc.
The equation is written properly. I'm not sure what you mean... the f^-1 is there.
7. (Original post by simon0)
I applied the reciprocal of dx/dy (so I did 1/(dx/dy) ). :-)
But that's what need to be proven.
8. Yes, it is correct in the last one you had there, but I was referring to the last line of your original...I was writing my response and by the time I had completed, you had made a correction to your initial post-which is the one (the first one) I was responding too. So now it is indeed correct.

(Original post by NotNotBatman)
The equation is written properly. I'm not sure what you mean... the f^-1 is there.
9. (Original post by luq_ali)
Yes, it is correct in the last one you had there, but I was referring to the last line of your original...I was writing my response and by the time I had completed, you had made a correction to your initial post-which is the one (the first one) I was responding too. So now it is indeed correct.
I haven't edited my post, maybe it took while to load or something I don't know.

But I still don't understand how my working is wrong :/

it isn't correct as it would mean that dy/dx at some point is always equal to it's reciprocal.
10. The equation as stated by Simon is correct, and that is what I was noting-you quoted it in responding to him and asking about it-what he gave was the correct one as far as I can tell.

(Original post by NotNotBatman)
I haven't edited my post, maybe it took while to load or something I don't know.

But I still don't understand how my working is wrong :/

it isn't correct as it would mean that dy/dx at some point is always equal to it's reciprocal.
11. (Original post by luq_ali)
The equation as stated by Simon is correct, and that is what I was noting-you quoted it in responding to him and asking about it-what he gave was the correct one as far as I can tell.
Oh ok, but what I want to know is how what I've written is incorrect (concerning the relationship between notation), because it seems like the answer was used to prove the original question.
12. One short way to show that dy/dx = 1/(dy/dx) is using the chain rule where:

y = f(x).

Differentiate with respect to y using the chain rule:

1 = d[f(x)] / dy = ( d[f(x)]/dx ) ( dx/dy) = (dy/dx)(dx/dy), as we defined y = f(x).

Therefore:

dy/dx = 1/( dx/dy ).

This is assuming dx/dy is not 0.

A more formal proof is using the inverse function theorem.
13. Your final equation in your last line-compare it what Simon has as his final line-and try to work back from there.

(Original post by NotNotBatman)
Oh ok, but what I want to know is how what I've written is incorrect (concerning the relationship between notation), because it seems like the answer was used to prove the original question.

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: October 29, 2017
Today on TSR

### He broke up with me because of long distance

Now I'm moving to his city

### University open days

Wed, 25 Jul '18
2. University of Buckingham
Wed, 25 Jul '18
3. Bournemouth University
Wed, 1 Aug '18
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams