Derivative of the inverse,so easy way of proving? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Proving that the derivative existsAlternative proof for differentiability of inverse function?If $f^-1$ has nowhere zero derivative, then $f$ is differentiableTrouble proving/disproving the existence of a particular derivative.one-sided continuity and one-sided derivative?Rudin's definition of derivativeCan't find mistake in proof that the derivative of a differentiable function is always continuous.Limit of a derivative of inverseTrying to understand discontinuities and domainsright derivative vs a right limit of the derivative times smallness factor
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Derivative of the inverse,so easy way of proving?
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Proving that the derivative existsAlternative proof for differentiability of inverse function?If $f^-1$ has nowhere zero derivative, then $f$ is differentiableTrouble proving/disproving the existence of a particular derivative.one-sided continuity and one-sided derivative?Rudin's definition of derivativeCan't find mistake in proof that the derivative of a differentiable function is always continuous.Limit of a derivative of inverseTrying to understand discontinuities and domainsright derivative vs a right limit of the derivative times smallness factor
$begingroup$
In Spivak :" Let $F$ be a continuous one one function defined on an interval and suppose that $F$ is differentiable at $F^-1(a) $ and $neq$ 0 so $F^-1$ is differentiable at $a$"
The limit $lim_h to 0 fracF^-1(a+h)-F^-1(a)h = lim_h to 0 fracF^-1(a+h)-F^-1(a)F(F^-1(a+h))-F(F^-1(a))$.
Since the inverse of this limit has the following face:$lim_h to 0 fracF(F^-1(a+h)-F(F^-1(a) F^-1(a+h)-F^-1(a)$.
This limit exist? Yeah since $F^-1$ is injective so $F^-1(a+h)-F^-1(a) neq 0$ for all $h$ except $=$ 0
And is continuous, so it exist and its value is $ F'(F^-1(a))$ and its inverse also exists, since I'm supposing it is different from $0$.
Is this right?
calculus
edited Apr 8 at 17:41
md2perpe
8,48411028
asked Apr 8 at 13:44
JordanJordan
185
New contributor
Jordan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
In Spivak :" Let $F$ be a continuous one one function defined on an interval and suppose that $F$ is differentiable at $F^-1(a) $ and $neq$ 0 so $F^-1$ is differentiable at $a$"
The limit $lim_h to 0 fracF^-1(a+h)-F^-1(a)h = lim_h to 0 fracF^-1(a+h)-F^-1(a)F(F^-1(a+h))-F(F^-1(a))$.
Since the inverse of this limit has the following face:$lim_h to 0 fracF(F^-1(a+h)-F(F^-1(a) F^-1(a+h)-F^-1(a)$.
This limit exist? Yeah since $F^-1$ is injective so $F^-1(a+h)-F^-1(a) neq 0$ for all $h$ except $=$ 0
And is continuous, so it exist and its value is $ F'(F^-1(a))$ and its inverse also exists, since I'm supposing it is different from $0$.
Is this right?
calculus
edited Apr 8 at 17:41
md2perpe
8,48411028
asked Apr 8 at 13:44
JordanJordan
185
New contributor
Jordan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
In Spivak :" Let $F$ be a continuous one one function defined on an interval and suppose that $F$ is differentiable at $F^-1(a) $ and $neq$ 0 so $F^-1$ is differentiable at $a$"
The limit $lim_h to 0 fracF^-1(a+h)-F^-1(a)h = lim_h to 0 fracF^-1(a+h)-F^-1(a)F(F^-1(a+h))-F(F^-1(a))$.
Since the inverse of this limit has the following face:$lim_h to 0 fracF(F^-1(a+h)-F(F^-1(a) F^-1(a+h)-F^-1(a)$.
This limit exist? Yeah since $F^-1$ is injective so $F^-1(a+h)-F^-1(a) neq 0$ for all $h$ except $=$ 0
And is continuous, so it exist and its value is $ F'(F^-1(a))$ and its inverse also exists, since I'm supposing it is different from $0$.
Is this right?
calculus
edited Apr 8 at 17:41
md2perpe
8,48411028
asked Apr 8 at 13:44
JordanJordan
185
New contributor
Jordan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
In Spivak :" Let $F$ be a continuous one one function defined on an interval and suppose that $F$ is differentiable at $F^-1(a) $ and $neq$ 0 so $F^-1$ is differentiable at $a$"
The limit $lim_h to 0 fracF^-1(a+h)-F^-1(a)h = lim_h to 0 fracF^-1(a+h)-F^-1(a)F(F^-1(a+h))-F(F^-1(a))$.
Since the inverse of this limit has the following face:$lim_h to 0 fracF(F^-1(a+h)-F(F^-1(a) F^-1(a+h)-F^-1(a)$.
This limit exist? Yeah since $F^-1$ is injective so $F^-1(a+h)-F^-1(a) neq 0$ for all $h$ except $=$ 0
And is continuous, so it exist and its value is $ F'(F^-1(a))$ and its inverse also exists, since I'm supposing it is different from $0$.
Is this right?
calculus
calculus
edited Apr 8 at 17:41
md2perpe
8,48411028
asked Apr 8 at 13:44
JordanJordan
185
New contributor
Jordan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 8 at 17:41
md2perpe
8,48411028
asked Apr 8 at 13:44
JordanJordan
185
New contributor
Jordan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 8 at 17:41
md2perpe
8,48411028
edited Apr 8 at 17:41
md2perpe
8,48411028
edited Apr 8 at 17:41
md2perpe
8,48411028
8,48411028
asked Apr 8 at 13:44
JordanJordan
185
New contributor
Jordan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 8 at 13:44
JordanJordan
185
asked Apr 8 at 13:44
JordanJordan
185
185
New contributor
Jordan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Jordan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Jordan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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