For $f(x) = int_1^x fracdtt$ prove that $f^-1(y) = fracddyf^-1(y)$

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I am looking for verification on proving that for $$
f(x) = int_1^x fracdtt
$$
the derivative of the inverse function is the inverse function itself: $$
f^-1(y) = fracddyf^-1(y).
$$

Obviously the statement is true because

$f(x) = log(x)$ and therefore $f^-1(y) = exp(y)$

And because $exp(y) = fracddyexp(y)$.

However, I want to prove the fact using only the definitions of $f$ and $f^-1$.

So far I know that by the fundamental theorem of calculus $$
f'(x) = 1/x.
$$

And by the inverse function theorem I also know that $$
fracddy f^-1(y) = frac1f'(x)~ text such that f(x)=y.
$$

And therefore $$
fracddy f^-1(y) = frac1f'(f^-1(y)) = f^-1(y).
$$

So I guess that solves it... ?

edited Jul 20 at 20:46

AOrtiz

8,86621238

asked Jul 20 at 20:16

ted

446312

add a commentÂ |Â

up vote
4
down vote

favorite

I am looking for verification on proving that for $$
f(x) = int_1^x fracdtt
$$
the derivative of the inverse function is the inverse function itself: $$
f^-1(y) = fracddyf^-1(y).
$$

Obviously the statement is true because

$f(x) = log(x)$ and therefore $f^-1(y) = exp(y)$

And because $exp(y) = fracddyexp(y)$.

However, I want to prove the fact using only the definitions of $f$ and $f^-1$.

So far I know that by the fundamental theorem of calculus $$
f'(x) = 1/x.
$$

And by the inverse function theorem I also know that $$
fracddy f^-1(y) = frac1f'(x)~ text such that f(x)=y.
$$

And therefore $$
fracddy f^-1(y) = frac1f'(f^-1(y)) = f^-1(y).
$$

So I guess that solves it... ?

edited Jul 20 at 20:46

AOrtiz

8,86621238

asked Jul 20 at 20:16

ted

446312

add a commentÂ |Â

up vote
4
down vote

favorite

I am looking for verification on proving that for $$
f(x) = int_1^x fracdtt
$$
the derivative of the inverse function is the inverse function itself: $$
f^-1(y) = fracddyf^-1(y).
$$

Obviously the statement is true because

$f(x) = log(x)$ and therefore $f^-1(y) = exp(y)$

And because $exp(y) = fracddyexp(y)$.

However, I want to prove the fact using only the definitions of $f$ and $f^-1$.

So far I know that by the fundamental theorem of calculus $$
f'(x) = 1/x.
$$

And by the inverse function theorem I also know that $$
fracddy f^-1(y) = frac1f'(x)~ text such that f(x)=y.
$$

And therefore $$
fracddy f^-1(y) = frac1f'(f^-1(y)) = f^-1(y).
$$

So I guess that solves it... ?

edited Jul 20 at 20:46

AOrtiz

8,86621238

asked Jul 20 at 20:16

ted

446312

I am looking for verification on proving that for $$
f(x) = int_1^x fracdtt
$$
the derivative of the inverse function is the inverse function itself: $$
f^-1(y) = fracddyf^-1(y).
$$

Obviously the statement is true because

$f(x) = log(x)$ and therefore $f^-1(y) = exp(y)$

And because $exp(y) = fracddyexp(y)$.

However, I want to prove the fact using only the definitions of $f$ and $f^-1$.

So far I know that by the fundamental theorem of calculus $$
f'(x) = 1/x.
$$

And by the inverse function theorem I also know that $$
fracddy f^-1(y) = frac1f'(x)~ text such that f(x)=y.
$$

And therefore $$
fracddy f^-1(y) = frac1f'(f^-1(y)) = f^-1(y).
$$

So I guess that solves it... ?

edited Jul 20 at 20:46

AOrtiz

8,86621238

asked Jul 20 at 20:16

ted

446312

edited Jul 20 at 20:46

AOrtiz

8,86621238

edited Jul 20 at 20:46

AOrtiz

8,86621238

edited Jul 20 at 20:46

AOrtiz

8,86621238

asked Jul 20 at 20:16

ted

446312

asked Jul 20 at 20:16

ted

446312

asked Jul 20 at 20:16

ted

446312

add a commentÂ |Â

1 Answer
1

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up vote
1
down vote

accepted

Your proof is mostly fine. To make it airtight, you should mention that the derivative of $f$ is strictly positive on $(0,infty)$ so the inverse function theorem actually applies.

answered Jul 20 at 20:34

AOrtiz

8,86621238

Yes, I did that elsewhere. Thanks!
â€“Â ted
Jul 20 at 20:45

add a commentÂ |Â

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active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
1
down vote

accepted

Your proof is mostly fine. To make it airtight, you should mention that the derivative of $f$ is strictly positive on $(0,infty)$ so the inverse function theorem actually applies.

answered Jul 20 at 20:34

AOrtiz

8,86621238

Yes, I did that elsewhere. Thanks!
â€“Â ted
Jul 20 at 20:45

add a commentÂ |Â

up vote
1
down vote

accepted

Your proof is mostly fine. To make it airtight, you should mention that the derivative of $f$ is strictly positive on $(0,infty)$ so the inverse function theorem actually applies.

answered Jul 20 at 20:34

AOrtiz

8,86621238

Yes, I did that elsewhere. Thanks!
â€“Â ted
Jul 20 at 20:45

add a commentÂ |Â

up vote
1
down vote

accepted

Your proof is mostly fine. To make it airtight, you should mention that the derivative of $f$ is strictly positive on $(0,infty)$ so the inverse function theorem actually applies.

answered Jul 20 at 20:34

AOrtiz

8,86621238

Your proof is mostly fine. To make it airtight, you should mention that the derivative of $f$ is strictly positive on $(0,infty)$ so the inverse function theorem actually applies.

answered Jul 20 at 20:34

AOrtiz

8,86621238

answered Jul 20 at 20:34

AOrtiz

8,86621238

answered Jul 20 at 20:34

AOrtiz

8,86621238

answered Jul 20 at 20:34

AOrtiz

8,86621238

Yes, I did that elsewhere. Thanks!
â€“Â ted
Jul 20 at 20:45

add a commentÂ |Â

Yes, I did that elsewhere. Thanks!
â€“Â ted
Jul 20 at 20:45

Yes, I did that elsewhere. Thanks!
â€“Â ted
Jul 20 at 20:45

add a commentÂ |Â

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