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Stuck on derivative of inverse function problem!
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I am super stuck on a problem. I am on the chapter of my calculus book on the derivatives on inverse functions but I can't figure it out.
[Note: I am writing the inverse of $f$ as $f^-1$]
The problem is:
Find $(f^-1)'(5)$ where $f(x) = 3x^3 + 4x^2 + 6x + 5$.
The back of the book says the answer is $1/6 $
I know that $(f^-1)'(a)$ = $1 / f'( f^1(a) )$
I used symlab to find $f(x) = 5$ and it gives me:
$-(2/3) * (+ or -) i(sqrt14 / 3)$
which is a complex number so that can't be what I'm looking for.
Then i tried implicit derivation to find $dy/dx$ and that just gave me:
$9x^2 + 8x + 6$
And plugging in $5$ gives a number way too big.
I have absolutely no idea what I'm doing wrong. I've watched all kinds of videos on this and reread the chapter a bunch of times and still can't get it. They always find the equation (f^-1). But if you just find that equation, then why not just take it's derivative rather than the weird [ $1 / f'( f^-1(x))$ ] ?
I must be missing some basic idea of this, but I don't know what.
I really appreciate any help! Thanks!
derivatives inverse-function
edited Jul 16 at 16:55
Robert Z
84.2k955123
asked Jul 16 at 16:44
Dr. Snow
523
add a comment |Â
up vote
0
down vote
favorite
I am super stuck on a problem. I am on the chapter of my calculus book on the derivatives on inverse functions but I can't figure it out.
[Note: I am writing the inverse of $f$ as $f^-1$]
The problem is:
Find $(f^-1)'(5)$ where $f(x) = 3x^3 + 4x^2 + 6x + 5$.
The back of the book says the answer is $1/6 $
I know that $(f^-1)'(a)$ = $1 / f'( f^1(a) )$
I used symlab to find $f(x) = 5$ and it gives me:
$-(2/3) * (+ or -) i(sqrt14 / 3)$
which is a complex number so that can't be what I'm looking for.
Then i tried implicit derivation to find $dy/dx$ and that just gave me:
$9x^2 + 8x + 6$
And plugging in $5$ gives a number way too big.
I have absolutely no idea what I'm doing wrong. I've watched all kinds of videos on this and reread the chapter a bunch of times and still can't get it. They always find the equation (f^-1). But if you just find that equation, then why not just take it's derivative rather than the weird [ $1 / f'( f^-1(x))$ ] ?
I must be missing some basic idea of this, but I don't know what.
I really appreciate any help! Thanks!
derivatives inverse-function
edited Jul 16 at 16:55
Robert Z
84.2k955123
asked Jul 16 at 16:44
Dr. Snow
523
The idea is that you cannot simply differentiate the inverse function if you can’t define it! For your example, solving for $x$ is possible, but problematic at best. For the equation you referenced, you need only $f^-1(5)$ and $f’(x)$. Note that $f(0)=5$.
– coreyman317
Jul 16 at 16:48
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am super stuck on a problem. I am on the chapter of my calculus book on the derivatives on inverse functions but I can't figure it out.
[Note: I am writing the inverse of $f$ as $f^-1$]
The problem is:
Find $(f^-1)'(5)$ where $f(x) = 3x^3 + 4x^2 + 6x + 5$.
The back of the book says the answer is $1/6 $
I know that $(f^-1)'(a)$ = $1 / f'( f^1(a) )$
I used symlab to find $f(x) = 5$ and it gives me:
$-(2/3) * (+ or -) i(sqrt14 / 3)$
which is a complex number so that can't be what I'm looking for.
Then i tried implicit derivation to find $dy/dx$ and that just gave me:
$9x^2 + 8x + 6$
And plugging in $5$ gives a number way too big.
I have absolutely no idea what I'm doing wrong. I've watched all kinds of videos on this and reread the chapter a bunch of times and still can't get it. They always find the equation (f^-1). But if you just find that equation, then why not just take it's derivative rather than the weird [ $1 / f'( f^-1(x))$ ] ?
I must be missing some basic idea of this, but I don't know what.
I really appreciate any help! Thanks!
derivatives inverse-function
edited Jul 16 at 16:55
Robert Z
84.2k955123
asked Jul 16 at 16:44
Dr. Snow
523
I am super stuck on a problem. I am on the chapter of my calculus book on the derivatives on inverse functions but I can't figure it out.
[Note: I am writing the inverse of $f$ as $f^-1$]
The problem is:
Find $(f^-1)'(5)$ where $f(x) = 3x^3 + 4x^2 + 6x + 5$.
The back of the book says the answer is $1/6 $
I know that $(f^-1)'(a)$ = $1 / f'( f^1(a) )$
I used symlab to find $f(x) = 5$ and it gives me:
$-(2/3) * (+ or -) i(sqrt14 / 3)$
which is a complex number so that can't be what I'm looking for.
Then i tried implicit derivation to find $dy/dx$ and that just gave me:
$9x^2 + 8x + 6$
And plugging in $5$ gives a number way too big.
I have absolutely no idea what I'm doing wrong. I've watched all kinds of videos on this and reread the chapter a bunch of times and still can't get it. They always find the equation (f^-1). But if you just find that equation, then why not just take it's derivative rather than the weird [ $1 / f'( f^-1(x))$ ] ?
I must be missing some basic idea of this, but I don't know what.
I really appreciate any help! Thanks!
derivatives inverse-function
edited Jul 16 at 16:55
Robert Z
84.2k955123
asked Jul 16 at 16:44
Dr. Snow
523
edited Jul 16 at 16:55
Robert Z
84.2k955123
edited Jul 16 at 16:55
Robert Z
84.2k955123
edited Jul 16 at 16:55
Robert Z
84.2k955123
84.2k955123
asked Jul 16 at 16:44
Dr. Snow
523
asked Jul 16 at 16:44
Dr. Snow
523
asked Jul 16 at 16:44
Dr. Snow
523
523
The idea is that you cannot simply differentiate the inverse function if you can’t define it! For your example, solving for $x$ is possible, but problematic at best. For the equation you referenced, you need only $f^-1(5)$ and $f’(x)$. Note that $f(0)=5$.
– coreyman317
Jul 16 at 16:48
add a comment |Â
The idea is that you cannot simply differentiate the inverse function if you can’t define it! For your example, solving for $x$ is possible, but problematic at best. For the equation you referenced, you need only $f^-1(5)$ and $f’(x)$. Note that $f(0)=5$.
– coreyman317
Jul 16 at 16:48
The idea is that you cannot simply differentiate the inverse function if you can’t define it! For your example, solving for $x$ is possible, but problematic at best. For the equation you referenced, you need only $f^-1(5)$ and $f’(x)$. Note that $f(0)=5$.
– coreyman317
Jul 16 at 16:48
The idea is that you cannot simply differentiate the inverse function if you can’t define it! For your example, solving for $x$ is possible, but problematic at best. For the equation you referenced, you need only $f^-1(5)$ and $f’(x)$. Note that $f(0)=5$.
– coreyman317
Jul 16 at 16:48
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
Note that the polynomial $f(x)=3x^3 + 4x^2 + 6x + 5$ is a strictly increasing function. Hence $f$ has an inverse and $f^-1:mathbbRto mathbbR$.
Now $f(0)=5$. What is $f'(0)$? Is this value related with derivative of $f^-1$ at $5$?
answered Jul 16 at 16:49
Robert Z
84.2k955123
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Note that the polynomial $f(x)=3x^3 + 4x^2 + 6x + 5$ is a strictly increasing function. Hence $f$ has an inverse and $f^-1:mathbbRto mathbbR$.
Now $f(0)=5$. What is $f'(0)$? Is this value related with derivative of $f^-1$ at $5$?
answered Jul 16 at 16:49
Robert Z
84.2k955123
add a comment |Â
up vote
3
down vote
accepted
Note that the polynomial $f(x)=3x^3 + 4x^2 + 6x + 5$ is a strictly increasing function. Hence $f$ has an inverse and $f^-1:mathbbRto mathbbR$.
Now $f(0)=5$. What is $f'(0)$? Is this value related with derivative of $f^-1$ at $5$?
answered Jul 16 at 16:49
Robert Z
84.2k955123
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Note that the polynomial $f(x)=3x^3 + 4x^2 + 6x + 5$ is a strictly increasing function. Hence $f$ has an inverse and $f^-1:mathbbRto mathbbR$.
Now $f(0)=5$. What is $f'(0)$? Is this value related with derivative of $f^-1$ at $5$?
answered Jul 16 at 16:49
Robert Z
84.2k955123
Note that the polynomial $f(x)=3x^3 + 4x^2 + 6x + 5$ is a strictly increasing function. Hence $f$ has an inverse and $f^-1:mathbbRto mathbbR$.
Now $f(0)=5$. What is $f'(0)$? Is this value related with derivative of $f^-1$ at $5$?
answered Jul 16 at 16:49
Robert Z
84.2k955123
edited Jul 16 at 17:04
answered Jul 16 at 16:49
Robert Z
84.2k955123
answered Jul 16 at 16:49
Robert Z
84.2k955123
answered Jul 16 at 16:49
Robert Z
84.2k955123
84.2k955123
add a comment |Â
add a comment |Â
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The idea is that you cannot simply differentiate the inverse function if you can’t define it! For your example, solving for $x$ is possible, but problematic at best. For the equation you referenced, you need only $f^-1(5)$ and $f’(x)$. Note that $f(0)=5$.
– coreyman317
Jul 16 at 16:48