How to show that the function $f(x) = sin x + tan x$ on $(-pi/2, pi/2)$ has an inverse












1












$begingroup$


Given that $f(x)=sin x+tan x$ on $(-pi/2,pi/2)$. Show that it has an inverse function. Find the derivative of the inverse of $f(x)$ at $x=0$.



What is the appropriate way to show that a function has an inverse?
How can I inverse two trigonometric function so I can derive them when they are tied up each other with sum symbol?






calculus algebra-precalculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Welcome to this site. What are the conditions necessary for a function to have an inverse? Heard of bijectivity? For the second part, use the inverse function theorem (which is basically chain rule).
    $endgroup$
    – Deepak
    Dec 8 '18 at 9:44






  • 2




    $begingroup$
    In order for a function to have an inverse, the only thing one need is the function is injective. If the function is injective, then it has an inverse over its range. Since both $sin x$ and $tan x$ is strictly increasing over $(-frac{pi}{2},frac{pi}{2})$, ....
    $endgroup$
    – achille hui
    Dec 8 '18 at 9:49










  • $begingroup$
    @achille hui I think that we need not only injection but bijection to have an inverse for a function.
    $endgroup$
    – hamza boulahia
    Dec 8 '18 at 10:01










  • $begingroup$
    @hamzaboulahia Injection is sufficient.
    $endgroup$
    – Rebellos
    Dec 8 '18 at 10:02










  • $begingroup$
    @hamzaboulahia you need bijection only when you want the domain of your inverse equals to the whole set that your function can take values.
    $endgroup$
    – achille hui
    Dec 8 '18 at 10:05
















1












$begingroup$


Given that $f(x)=sin x+tan x$ on $(-pi/2,pi/2)$. Show that it has an inverse function. Find the derivative of the inverse of $f(x)$ at $x=0$.



What is the appropriate way to show that a function has an inverse?
How can I inverse two trigonometric function so I can derive them when they are tied up each other with sum symbol?






calculus algebra-precalculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Welcome to this site. What are the conditions necessary for a function to have an inverse? Heard of bijectivity? For the second part, use the inverse function theorem (which is basically chain rule).
    $endgroup$
    – Deepak
    Dec 8 '18 at 9:44






  • 2




    $begingroup$
    In order for a function to have an inverse, the only thing one need is the function is injective. If the function is injective, then it has an inverse over its range. Since both $sin x$ and $tan x$ is strictly increasing over $(-frac{pi}{2},frac{pi}{2})$, ....
    $endgroup$
    – achille hui
    Dec 8 '18 at 9:49










  • $begingroup$
    @achille hui I think that we need not only injection but bijection to have an inverse for a function.
    $endgroup$
    – hamza boulahia
    Dec 8 '18 at 10:01










  • $begingroup$
    @hamzaboulahia Injection is sufficient.
    $endgroup$
    – Rebellos
    Dec 8 '18 at 10:02










  • $begingroup$
    @hamzaboulahia you need bijection only when you want the domain of your inverse equals to the whole set that your function can take values.
    $endgroup$
    – achille hui
    Dec 8 '18 at 10:05














1












1








1





$begingroup$


Given that $f(x)=sin x+tan x$ on $(-pi/2,pi/2)$. Show that it has an inverse function. Find the derivative of the inverse of $f(x)$ at $x=0$.



What is the appropriate way to show that a function has an inverse?
How can I inverse two trigonometric function so I can derive them when they are tied up each other with sum symbol?






calculus algebra-precalculus







share|cite|improve this question











$endgroup$




Given that $f(x)=sin x+tan x$ on $(-pi/2,pi/2)$. Show that it has an inverse function. Find the derivative of the inverse of $f(x)$ at $x=0$.



What is the appropriate way to show that a function has an inverse?
How can I inverse two trigonometric function so I can derive them when they are tied up each other with sum symbol?






calculus algebra-precalculus




calculus algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 11:03









N. F. Taussig

44.1k93356




44.1k93356










asked Dec 8 '18 at 9:40









HouseBTHouseBT

154




154








  • 1




    $begingroup$
    Welcome to this site. What are the conditions necessary for a function to have an inverse? Heard of bijectivity? For the second part, use the inverse function theorem (which is basically chain rule).
    $endgroup$
    – Deepak
    Dec 8 '18 at 9:44






  • 2




    $begingroup$
    In order for a function to have an inverse, the only thing one need is the function is injective. If the function is injective, then it has an inverse over its range. Since both $sin x$ and $tan x$ is strictly increasing over $(-frac{pi}{2},frac{pi}{2})$, ....
    $endgroup$
    – achille hui
    Dec 8 '18 at 9:49










  • $begingroup$
    @achille hui I think that we need not only injection but bijection to have an inverse for a function.
    $endgroup$
    – hamza boulahia
    Dec 8 '18 at 10:01










  • $begingroup$
    @hamzaboulahia Injection is sufficient.
    $endgroup$
    – Rebellos
    Dec 8 '18 at 10:02










  • $begingroup$
    @hamzaboulahia you need bijection only when you want the domain of your inverse equals to the whole set that your function can take values.
    $endgroup$
    – achille hui
    Dec 8 '18 at 10:05














  • 1




    $begingroup$
    Welcome to this site. What are the conditions necessary for a function to have an inverse? Heard of bijectivity? For the second part, use the inverse function theorem (which is basically chain rule).
    $endgroup$
    – Deepak
    Dec 8 '18 at 9:44






  • 2




    $begingroup$
    In order for a function to have an inverse, the only thing one need is the function is injective. If the function is injective, then it has an inverse over its range. Since both $sin x$ and $tan x$ is strictly increasing over $(-frac{pi}{2},frac{pi}{2})$, ....
    $endgroup$
    – achille hui
    Dec 8 '18 at 9:49










  • $begingroup$
    @achille hui I think that we need not only injection but bijection to have an inverse for a function.
    $endgroup$
    – hamza boulahia
    Dec 8 '18 at 10:01










  • $begingroup$
    @hamzaboulahia Injection is sufficient.
    $endgroup$
    – Rebellos
    Dec 8 '18 at 10:02










  • $begingroup$
    @hamzaboulahia you need bijection only when you want the domain of your inverse equals to the whole set that your function can take values.
    $endgroup$
    – achille hui
    Dec 8 '18 at 10:05








1




1




$begingroup$
Welcome to this site. What are the conditions necessary for a function to have an inverse? Heard of bijectivity? For the second part, use the inverse function theorem (which is basically chain rule).
$endgroup$
– Deepak
Dec 8 '18 at 9:44




$begingroup$
Welcome to this site. What are the conditions necessary for a function to have an inverse? Heard of bijectivity? For the second part, use the inverse function theorem (which is basically chain rule).
$endgroup$
– Deepak
Dec 8 '18 at 9:44




2




2




$begingroup$
In order for a function to have an inverse, the only thing one need is the function is injective. If the function is injective, then it has an inverse over its range. Since both $sin x$ and $tan x$ is strictly increasing over $(-frac{pi}{2},frac{pi}{2})$, ....
$endgroup$
– achille hui
Dec 8 '18 at 9:49




$begingroup$
In order for a function to have an inverse, the only thing one need is the function is injective. If the function is injective, then it has an inverse over its range. Since both $sin x$ and $tan x$ is strictly increasing over $(-frac{pi}{2},frac{pi}{2})$, ....
$endgroup$
– achille hui
Dec 8 '18 at 9:49












$begingroup$
@achille hui I think that we need not only injection but bijection to have an inverse for a function.
$endgroup$
– hamza boulahia
Dec 8 '18 at 10:01




$begingroup$
@achille hui I think that we need not only injection but bijection to have an inverse for a function.
$endgroup$
– hamza boulahia
Dec 8 '18 at 10:01












$begingroup$
@hamzaboulahia Injection is sufficient.
$endgroup$
– Rebellos
Dec 8 '18 at 10:02




$begingroup$
@hamzaboulahia Injection is sufficient.
$endgroup$
– Rebellos
Dec 8 '18 at 10:02












$begingroup$
@hamzaboulahia you need bijection only when you want the domain of your inverse equals to the whole set that your function can take values.
$endgroup$
– achille hui
Dec 8 '18 at 10:05




$begingroup$
@hamzaboulahia you need bijection only when you want the domain of your inverse equals to the whole set that your function can take values.
$endgroup$
– achille hui
Dec 8 '18 at 10:05










2 Answers
2






active

oldest

votes


















1












$begingroup$

In order to show that this function has an inverse, you need to prove that it's "$1-1$", thus meaning that the given functional (function in our simple case) is injective.



But, $sin x$ and $tan x$ are strictly increasing over $left( -frac{pi}{2}, frac{pi}{2}right)$. This means that $f(x)$ is strictly increasing which implies that $f(x)$ is also "$1-1"$ $implies$ $f$ can be inversed.



Now, the expression of $f$ cannot be inversed explicitly (or to put it better, it can be, but it's impossible to be carried out by hand).



But, consider the following standard Theorem proved in pre-calculus courses :




Theorem : Suppose that $f$ has an inverse function $f^{-1}$. If $f$ is differentiable at $f^{-1}$ and $f '(f^{-1}(x))$ is not equal to zero, then $f^{-1}$ is differentiable at $x$ and the following differentiation formula holds :
$$(f^{-1})'(x)=frac{1}{f'(f^{-1}(x))}$$
Proof :



Let $y_0in mathbb{R}$. Then $exists x_0inmathbb{R} : f(x_0)=y_0$. Then $xto x_0 implies y=f(x)to f(x_0)=y_0$.
By definition, it is :
$$(f^{-1})'(y_0)=lim_{yrightarrow y_0}frac{f^{-1}(y)-f^{-1}(y_0)}{y-y_0}=lim_{yrightarrow y_0}frac{f^{-1}(f(x))-f^{-1}(f(x_0))}{f(x)-f(x_0)}\=lim_{xrightarrow x_0}frac{x-x_0}{f(x)-f(x_0)}=frac{1}{f'(x_0)}=frac{1}{f'(f^{-1}(y_0))}$$




Can you now use that theorem to find the derivative of the inverse at $x=0$ ?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Thank you for showing me how to prove that if a function has an inverse. So, Here is my solution; The inverse of it is x=siny+tany and x is 0. So -tany=siny, therefore cosy=-1 and y is π at the given point. The only thing left is to derive the f(x) and put π in it right?
    $endgroup$
    – HouseBT
    Dec 8 '18 at 10:15












  • $begingroup$
    Your elaboration in terms of explaining is very complicated and hard to follow. Maybe you would like to typeset using MathJax? By the way if an answer was helpful you may use the vote up button and the answer that fits you the best may be marked as accepted using the tick marker.
    $endgroup$
    – Rebellos
    Dec 8 '18 at 10:22










  • $begingroup$
    I do not know how to use a typeset yet. What i did was just inversing f(x) f(x)=y we can describe it like that. To inverse i just changed x's with y's so function becomes like this one below: x=siny+tany, due to the question we are at the point x=0, so equation becomes 0=siny+tany. After solving it I figure that y is equal to π at the point x=0. Finally I used the theorem to solve it. Did I do it right? If not, Where was i wrong?
    $endgroup$
    – HouseBT
    Dec 8 '18 at 10:28












  • $begingroup$
    What you did is wrong. For $x=0$, it is : $f(0) = 0$. This means that $f^{-1}(0) = 0$. Can you now continue to calculate $(f^{-1})'(0)$ ?
    $endgroup$
    – Rebellos
    Dec 8 '18 at 10:30










  • $begingroup$
    Okay now i understand my mistake and the right path to follow thank you for your time ^^
    $endgroup$
    – HouseBT
    Dec 8 '18 at 10:37





















1












$begingroup$

$star$For a real function such as the given function $f$. To show that $d$ has an inverse function, you have to show that $f$ is strictly monotonic on the given interval $]-frac{pi}{2},frac{pi}{2}[$.



$star$ For the second question, you don't have to calculate the inverse the each function to find the inverse of of $f$ because; If $f=g+h$ then, $$ f^{-1}neq g^{-1}+h^{-1}$$
But then again, you are only asked to calculate the inverse at a certain point $x=0$. So you can use the formula $$ (f^{-1})'(x)=frac{1}{f'(f^{-1}(x))}$$






share|cite|improve this answer









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3030894%2fhow-to-show-that-the-function-fx-sin-x-tan-x-on-pi-2-pi-2-has%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    In order to show that this function has an inverse, you need to prove that it's "$1-1$", thus meaning that the given functional (function in our simple case) is injective.



    But, $sin x$ and $tan x$ are strictly increasing over $left( -frac{pi}{2}, frac{pi}{2}right)$. This means that $f(x)$ is strictly increasing which implies that $f(x)$ is also "$1-1"$ $implies$ $f$ can be inversed.



    Now, the expression of $f$ cannot be inversed explicitly (or to put it better, it can be, but it's impossible to be carried out by hand).



    But, consider the following standard Theorem proved in pre-calculus courses :




    Theorem : Suppose that $f$ has an inverse function $f^{-1}$. If $f$ is differentiable at $f^{-1}$ and $f '(f^{-1}(x))$ is not equal to zero, then $f^{-1}$ is differentiable at $x$ and the following differentiation formula holds :
    $$(f^{-1})'(x)=frac{1}{f'(f^{-1}(x))}$$
    Proof :



    Let $y_0in mathbb{R}$. Then $exists x_0inmathbb{R} : f(x_0)=y_0$. Then $xto x_0 implies y=f(x)to f(x_0)=y_0$.
    By definition, it is :
    $$(f^{-1})'(y_0)=lim_{yrightarrow y_0}frac{f^{-1}(y)-f^{-1}(y_0)}{y-y_0}=lim_{yrightarrow y_0}frac{f^{-1}(f(x))-f^{-1}(f(x_0))}{f(x)-f(x_0)}\=lim_{xrightarrow x_0}frac{x-x_0}{f(x)-f(x_0)}=frac{1}{f'(x_0)}=frac{1}{f'(f^{-1}(y_0))}$$




    Can you now use that theorem to find the derivative of the inverse at $x=0$ ?






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you for showing me how to prove that if a function has an inverse. So, Here is my solution; The inverse of it is x=siny+tany and x is 0. So -tany=siny, therefore cosy=-1 and y is π at the given point. The only thing left is to derive the f(x) and put π in it right?
      $endgroup$
      – HouseBT
      Dec 8 '18 at 10:15












    • $begingroup$
      Your elaboration in terms of explaining is very complicated and hard to follow. Maybe you would like to typeset using MathJax? By the way if an answer was helpful you may use the vote up button and the answer that fits you the best may be marked as accepted using the tick marker.
      $endgroup$
      – Rebellos
      Dec 8 '18 at 10:22










    • $begingroup$
      I do not know how to use a typeset yet. What i did was just inversing f(x) f(x)=y we can describe it like that. To inverse i just changed x's with y's so function becomes like this one below: x=siny+tany, due to the question we are at the point x=0, so equation becomes 0=siny+tany. After solving it I figure that y is equal to π at the point x=0. Finally I used the theorem to solve it. Did I do it right? If not, Where was i wrong?
      $endgroup$
      – HouseBT
      Dec 8 '18 at 10:28












    • $begingroup$
      What you did is wrong. For $x=0$, it is : $f(0) = 0$. This means that $f^{-1}(0) = 0$. Can you now continue to calculate $(f^{-1})'(0)$ ?
      $endgroup$
      – Rebellos
      Dec 8 '18 at 10:30










    • $begingroup$
      Okay now i understand my mistake and the right path to follow thank you for your time ^^
      $endgroup$
      – HouseBT
      Dec 8 '18 at 10:37


















    1












    $begingroup$

    In order to show that this function has an inverse, you need to prove that it's "$1-1$", thus meaning that the given functional (function in our simple case) is injective.



    But, $sin x$ and $tan x$ are strictly increasing over $left( -frac{pi}{2}, frac{pi}{2}right)$. This means that $f(x)$ is strictly increasing which implies that $f(x)$ is also "$1-1"$ $implies$ $f$ can be inversed.



    Now, the expression of $f$ cannot be inversed explicitly (or to put it better, it can be, but it's impossible to be carried out by hand).



    But, consider the following standard Theorem proved in pre-calculus courses :




    Theorem : Suppose that $f$ has an inverse function $f^{-1}$. If $f$ is differentiable at $f^{-1}$ and $f '(f^{-1}(x))$ is not equal to zero, then $f^{-1}$ is differentiable at $x$ and the following differentiation formula holds :
    $$(f^{-1})'(x)=frac{1}{f'(f^{-1}(x))}$$
    Proof :



    Let $y_0in mathbb{R}$. Then $exists x_0inmathbb{R} : f(x_0)=y_0$. Then $xto x_0 implies y=f(x)to f(x_0)=y_0$.
    By definition, it is :
    $$(f^{-1})'(y_0)=lim_{yrightarrow y_0}frac{f^{-1}(y)-f^{-1}(y_0)}{y-y_0}=lim_{yrightarrow y_0}frac{f^{-1}(f(x))-f^{-1}(f(x_0))}{f(x)-f(x_0)}\=lim_{xrightarrow x_0}frac{x-x_0}{f(x)-f(x_0)}=frac{1}{f'(x_0)}=frac{1}{f'(f^{-1}(y_0))}$$




    Can you now use that theorem to find the derivative of the inverse at $x=0$ ?






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thank you for showing me how to prove that if a function has an inverse. So, Here is my solution; The inverse of it is x=siny+tany and x is 0. So -tany=siny, therefore cosy=-1 and y is π at the given point. The only thing left is to derive the f(x) and put π in it right?
      $endgroup$
      – HouseBT
      Dec 8 '18 at 10:15












    • $begingroup$
      Your elaboration in terms of explaining is very complicated and hard to follow. Maybe you would like to typeset using MathJax? By the way if an answer was helpful you may use the vote up button and the answer that fits you the best may be marked as accepted using the tick marker.
      $endgroup$
      – Rebellos
      Dec 8 '18 at 10:22










    • $begingroup$
      I do not know how to use a typeset yet. What i did was just inversing f(x) f(x)=y we can describe it like that. To inverse i just changed x's with y's so function becomes like this one below: x=siny+tany, due to the question we are at the point x=0, so equation becomes 0=siny+tany. After solving it I figure that y is equal to π at the point x=0. Finally I used the theorem to solve it. Did I do it right? If not, Where was i wrong?
      $endgroup$
      – HouseBT
      Dec 8 '18 at 10:28












    • $begingroup$
      What you did is wrong. For $x=0$, it is : $f(0) = 0$. This means that $f^{-1}(0) = 0$. Can you now continue to calculate $(f^{-1})'(0)$ ?
      $endgroup$
      – Rebellos
      Dec 8 '18 at 10:30










    • $begingroup$
      Okay now i understand my mistake and the right path to follow thank you for your time ^^
      $endgroup$
      – HouseBT
      Dec 8 '18 at 10:37
















    1












    1








    1





    $begingroup$

    In order to show that this function has an inverse, you need to prove that it's "$1-1$", thus meaning that the given functional (function in our simple case) is injective.



    But, $sin x$ and $tan x$ are strictly increasing over $left( -frac{pi}{2}, frac{pi}{2}right)$. This means that $f(x)$ is strictly increasing which implies that $f(x)$ is also "$1-1"$ $implies$ $f$ can be inversed.



    Now, the expression of $f$ cannot be inversed explicitly (or to put it better, it can be, but it's impossible to be carried out by hand).



    But, consider the following standard Theorem proved in pre-calculus courses :




    Theorem : Suppose that $f$ has an inverse function $f^{-1}$. If $f$ is differentiable at $f^{-1}$ and $f '(f^{-1}(x))$ is not equal to zero, then $f^{-1}$ is differentiable at $x$ and the following differentiation formula holds :
    $$(f^{-1})'(x)=frac{1}{f'(f^{-1}(x))}$$
    Proof :



    Let $y_0in mathbb{R}$. Then $exists x_0inmathbb{R} : f(x_0)=y_0$. Then $xto x_0 implies y=f(x)to f(x_0)=y_0$.
    By definition, it is :
    $$(f^{-1})'(y_0)=lim_{yrightarrow y_0}frac{f^{-1}(y)-f^{-1}(y_0)}{y-y_0}=lim_{yrightarrow y_0}frac{f^{-1}(f(x))-f^{-1}(f(x_0))}{f(x)-f(x_0)}\=lim_{xrightarrow x_0}frac{x-x_0}{f(x)-f(x_0)}=frac{1}{f'(x_0)}=frac{1}{f'(f^{-1}(y_0))}$$




    Can you now use that theorem to find the derivative of the inverse at $x=0$ ?






    share|cite|improve this answer











    $endgroup$



    In order to show that this function has an inverse, you need to prove that it's "$1-1$", thus meaning that the given functional (function in our simple case) is injective.



    But, $sin x$ and $tan x$ are strictly increasing over $left( -frac{pi}{2}, frac{pi}{2}right)$. This means that $f(x)$ is strictly increasing which implies that $f(x)$ is also "$1-1"$ $implies$ $f$ can be inversed.



    Now, the expression of $f$ cannot be inversed explicitly (or to put it better, it can be, but it's impossible to be carried out by hand).



    But, consider the following standard Theorem proved in pre-calculus courses :




    Theorem : Suppose that $f$ has an inverse function $f^{-1}$. If $f$ is differentiable at $f^{-1}$ and $f '(f^{-1}(x))$ is not equal to zero, then $f^{-1}$ is differentiable at $x$ and the following differentiation formula holds :
    $$(f^{-1})'(x)=frac{1}{f'(f^{-1}(x))}$$
    Proof :



    Let $y_0in mathbb{R}$. Then $exists x_0inmathbb{R} : f(x_0)=y_0$. Then $xto x_0 implies y=f(x)to f(x_0)=y_0$.
    By definition, it is :
    $$(f^{-1})'(y_0)=lim_{yrightarrow y_0}frac{f^{-1}(y)-f^{-1}(y_0)}{y-y_0}=lim_{yrightarrow y_0}frac{f^{-1}(f(x))-f^{-1}(f(x_0))}{f(x)-f(x_0)}\=lim_{xrightarrow x_0}frac{x-x_0}{f(x)-f(x_0)}=frac{1}{f'(x_0)}=frac{1}{f'(f^{-1}(y_0))}$$




    Can you now use that theorem to find the derivative of the inverse at $x=0$ ?







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 8 '18 at 10:08

























    answered Dec 8 '18 at 10:00









    RebellosRebellos

    14.6k31247




    14.6k31247












    • $begingroup$
      Thank you for showing me how to prove that if a function has an inverse. So, Here is my solution; The inverse of it is x=siny+tany and x is 0. So -tany=siny, therefore cosy=-1 and y is π at the given point. The only thing left is to derive the f(x) and put π in it right?
      $endgroup$
      – HouseBT
      Dec 8 '18 at 10:15












    • $begingroup$
      Your elaboration in terms of explaining is very complicated and hard to follow. Maybe you would like to typeset using MathJax? By the way if an answer was helpful you may use the vote up button and the answer that fits you the best may be marked as accepted using the tick marker.
      $endgroup$
      – Rebellos
      Dec 8 '18 at 10:22










    • $begingroup$
      I do not know how to use a typeset yet. What i did was just inversing f(x) f(x)=y we can describe it like that. To inverse i just changed x's with y's so function becomes like this one below: x=siny+tany, due to the question we are at the point x=0, so equation becomes 0=siny+tany. After solving it I figure that y is equal to π at the point x=0. Finally I used the theorem to solve it. Did I do it right? If not, Where was i wrong?
      $endgroup$
      – HouseBT
      Dec 8 '18 at 10:28












    • $begingroup$
      What you did is wrong. For $x=0$, it is : $f(0) = 0$. This means that $f^{-1}(0) = 0$. Can you now continue to calculate $(f^{-1})'(0)$ ?
      $endgroup$
      – Rebellos
      Dec 8 '18 at 10:30










    • $begingroup$
      Okay now i understand my mistake and the right path to follow thank you for your time ^^
      $endgroup$
      – HouseBT
      Dec 8 '18 at 10:37




















    • $begingroup$
      Thank you for showing me how to prove that if a function has an inverse. So, Here is my solution; The inverse of it is x=siny+tany and x is 0. So -tany=siny, therefore cosy=-1 and y is π at the given point. The only thing left is to derive the f(x) and put π in it right?
      $endgroup$
      – HouseBT
      Dec 8 '18 at 10:15












    • $begingroup$
      Your elaboration in terms of explaining is very complicated and hard to follow. Maybe you would like to typeset using MathJax? By the way if an answer was helpful you may use the vote up button and the answer that fits you the best may be marked as accepted using the tick marker.
      $endgroup$
      – Rebellos
      Dec 8 '18 at 10:22










    • $begingroup$
      I do not know how to use a typeset yet. What i did was just inversing f(x) f(x)=y we can describe it like that. To inverse i just changed x's with y's so function becomes like this one below: x=siny+tany, due to the question we are at the point x=0, so equation becomes 0=siny+tany. After solving it I figure that y is equal to π at the point x=0. Finally I used the theorem to solve it. Did I do it right? If not, Where was i wrong?
      $endgroup$
      – HouseBT
      Dec 8 '18 at 10:28












    • $begingroup$
      What you did is wrong. For $x=0$, it is : $f(0) = 0$. This means that $f^{-1}(0) = 0$. Can you now continue to calculate $(f^{-1})'(0)$ ?
      $endgroup$
      – Rebellos
      Dec 8 '18 at 10:30










    • $begingroup$
      Okay now i understand my mistake and the right path to follow thank you for your time ^^
      $endgroup$
      – HouseBT
      Dec 8 '18 at 10:37


















    $begingroup$
    Thank you for showing me how to prove that if a function has an inverse. So, Here is my solution; The inverse of it is x=siny+tany and x is 0. So -tany=siny, therefore cosy=-1 and y is π at the given point. The only thing left is to derive the f(x) and put π in it right?
    $endgroup$
    – HouseBT
    Dec 8 '18 at 10:15






    $begingroup$
    Thank you for showing me how to prove that if a function has an inverse. So, Here is my solution; The inverse of it is x=siny+tany and x is 0. So -tany=siny, therefore cosy=-1 and y is π at the given point. The only thing left is to derive the f(x) and put π in it right?
    $endgroup$
    – HouseBT
    Dec 8 '18 at 10:15














    $begingroup$
    Your elaboration in terms of explaining is very complicated and hard to follow. Maybe you would like to typeset using MathJax? By the way if an answer was helpful you may use the vote up button and the answer that fits you the best may be marked as accepted using the tick marker.
    $endgroup$
    – Rebellos
    Dec 8 '18 at 10:22




    $begingroup$
    Your elaboration in terms of explaining is very complicated and hard to follow. Maybe you would like to typeset using MathJax? By the way if an answer was helpful you may use the vote up button and the answer that fits you the best may be marked as accepted using the tick marker.
    $endgroup$
    – Rebellos
    Dec 8 '18 at 10:22












    $begingroup$
    I do not know how to use a typeset yet. What i did was just inversing f(x) f(x)=y we can describe it like that. To inverse i just changed x's with y's so function becomes like this one below: x=siny+tany, due to the question we are at the point x=0, so equation becomes 0=siny+tany. After solving it I figure that y is equal to π at the point x=0. Finally I used the theorem to solve it. Did I do it right? If not, Where was i wrong?
    $endgroup$
    – HouseBT
    Dec 8 '18 at 10:28






    $begingroup$
    I do not know how to use a typeset yet. What i did was just inversing f(x) f(x)=y we can describe it like that. To inverse i just changed x's with y's so function becomes like this one below: x=siny+tany, due to the question we are at the point x=0, so equation becomes 0=siny+tany. After solving it I figure that y is equal to π at the point x=0. Finally I used the theorem to solve it. Did I do it right? If not, Where was i wrong?
    $endgroup$
    – HouseBT
    Dec 8 '18 at 10:28














    $begingroup$
    What you did is wrong. For $x=0$, it is : $f(0) = 0$. This means that $f^{-1}(0) = 0$. Can you now continue to calculate $(f^{-1})'(0)$ ?
    $endgroup$
    – Rebellos
    Dec 8 '18 at 10:30




    $begingroup$
    What you did is wrong. For $x=0$, it is : $f(0) = 0$. This means that $f^{-1}(0) = 0$. Can you now continue to calculate $(f^{-1})'(0)$ ?
    $endgroup$
    – Rebellos
    Dec 8 '18 at 10:30












    $begingroup$
    Okay now i understand my mistake and the right path to follow thank you for your time ^^
    $endgroup$
    – HouseBT
    Dec 8 '18 at 10:37






    $begingroup$
    Okay now i understand my mistake and the right path to follow thank you for your time ^^
    $endgroup$
    – HouseBT
    Dec 8 '18 at 10:37













    1












    $begingroup$

    $star$For a real function such as the given function $f$. To show that $d$ has an inverse function, you have to show that $f$ is strictly monotonic on the given interval $]-frac{pi}{2},frac{pi}{2}[$.



    $star$ For the second question, you don't have to calculate the inverse the each function to find the inverse of of $f$ because; If $f=g+h$ then, $$ f^{-1}neq g^{-1}+h^{-1}$$
    But then again, you are only asked to calculate the inverse at a certain point $x=0$. So you can use the formula $$ (f^{-1})'(x)=frac{1}{f'(f^{-1}(x))}$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      $star$For a real function such as the given function $f$. To show that $d$ has an inverse function, you have to show that $f$ is strictly monotonic on the given interval $]-frac{pi}{2},frac{pi}{2}[$.



      $star$ For the second question, you don't have to calculate the inverse the each function to find the inverse of of $f$ because; If $f=g+h$ then, $$ f^{-1}neq g^{-1}+h^{-1}$$
      But then again, you are only asked to calculate the inverse at a certain point $x=0$. So you can use the formula $$ (f^{-1})'(x)=frac{1}{f'(f^{-1}(x))}$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        $star$For a real function such as the given function $f$. To show that $d$ has an inverse function, you have to show that $f$ is strictly monotonic on the given interval $]-frac{pi}{2},frac{pi}{2}[$.



        $star$ For the second question, you don't have to calculate the inverse the each function to find the inverse of of $f$ because; If $f=g+h$ then, $$ f^{-1}neq g^{-1}+h^{-1}$$
        But then again, you are only asked to calculate the inverse at a certain point $x=0$. So you can use the formula $$ (f^{-1})'(x)=frac{1}{f'(f^{-1}(x))}$$






        share|cite|improve this answer









        $endgroup$



        $star$For a real function such as the given function $f$. To show that $d$ has an inverse function, you have to show that $f$ is strictly monotonic on the given interval $]-frac{pi}{2},frac{pi}{2}[$.



        $star$ For the second question, you don't have to calculate the inverse the each function to find the inverse of of $f$ because; If $f=g+h$ then, $$ f^{-1}neq g^{-1}+h^{-1}$$
        But then again, you are only asked to calculate the inverse at a certain point $x=0$. So you can use the formula $$ (f^{-1})'(x)=frac{1}{f'(f^{-1}(x))}$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 8 '18 at 9:59









        hamza boulahiahamza boulahia

        989419




        989419






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3030894%2fhow-to-show-that-the-function-fx-sin-x-tan-x-on-pi-2-pi-2-has%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Bundesstraße 106

            Image
            Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück
            Read more

            Verónica Boquete

            Image
            Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl
            Read more

            Ida-Boy-Ed-Garten

            Image
            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich
            Read more