Given $f(x) = arcsinleft(frac{2x}{1+x^2}right)$, find $f'(1)$.












1












$begingroup$



Let
$$
f(x) = arcsinleft(frac{2x}{1+x^2}right).
$$
What is the value of $f'(1)$?




The function splits into
$$
f(x) = begin{cases}
phantom{pi-,}2arctan(x), & text{if $-1 leq x leq 1$},\
pi - 2arctan(x), & text{if $x > 1$}.
end{cases}
$$



I differentiated the function corresponding to $1$ but my answer is coming to be $1$, but the book tells me the answer does not exist. How can we say the derivative doesn't exist?






derivatives trigonometry







share|cite|improve this question











$endgroup$

















    1












    $begingroup$



    Let
    $$
    f(x) = arcsinleft(frac{2x}{1+x^2}right).
    $$
    What is the value of $f'(1)$?




    The function splits into
    $$
    f(x) = begin{cases}
    phantom{pi-,}2arctan(x), & text{if $-1 leq x leq 1$},\
    pi - 2arctan(x), & text{if $x > 1$}.
    end{cases}
    $$



    I differentiated the function corresponding to $1$ but my answer is coming to be $1$, but the book tells me the answer does not exist. How can we say the derivative doesn't exist?






    derivatives trigonometry







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$



      Let
      $$
      f(x) = arcsinleft(frac{2x}{1+x^2}right).
      $$
      What is the value of $f'(1)$?




      The function splits into
      $$
      f(x) = begin{cases}
      phantom{pi-,}2arctan(x), & text{if $-1 leq x leq 1$},\
      pi - 2arctan(x), & text{if $x > 1$}.
      end{cases}
      $$



      I differentiated the function corresponding to $1$ but my answer is coming to be $1$, but the book tells me the answer does not exist. How can we say the derivative doesn't exist?






      derivatives trigonometry







      share|cite|improve this question











      $endgroup$





      Let
      $$
      f(x) = arcsinleft(frac{2x}{1+x^2}right).
      $$
      What is the value of $f'(1)$?




      The function splits into
      $$
      f(x) = begin{cases}
      phantom{pi-,}2arctan(x), & text{if $-1 leq x leq 1$},\
      pi - 2arctan(x), & text{if $x > 1$}.
      end{cases}
      $$



      I differentiated the function corresponding to $1$ but my answer is coming to be $1$, but the book tells me the answer does not exist. How can we say the derivative doesn't exist?






      derivatives trigonometry




      derivatives trigonometry






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 19 at 19:41









      Blue

      48.7k870156




      48.7k870156










      asked Aug 28 '18 at 14:23









      GENESECT GENESECT

      718




      718






















          3 Answers
          3






          active

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          1












          $begingroup$

          Note that $arcsin$ is differentiable in $(-1,1)$, but it is not differentiable at $pm1$. Since $frac{2times1}{1+1^2}=1$, it would be strange indeed if $f(x)=arcsinleft(frac{2x}{1+x^2}right)$ was differentiable at $1$.






          share|cite|improve this answer









          $endgroup$









          • 1




            $begingroup$
            What makes the function non differentiable at the extreme points.... Is there no neighbourhood to check verify slope of tangent at the extreme points?
            $endgroup$
            – GENESECT
            Aug 28 '18 at 14:29








          • 1




            $begingroup$
            The tangent lines to the graph of $arcsin$ at $pmleft(1,fracpi2right)$ are vertical lines. Therefore, $arcsin$ is not differentiable at $pm1$.
            $endgroup$
            – José Carlos Santos
            Aug 28 '18 at 14:31





















          1












          $begingroup$

          Notice that
          begin{align*}
          f'(x) &= frac{1}{sqrt{1-left(frac{2x}{1+x^2}right)^2}}frac{d}{dx}left(frac{2x}{1+x^2}right)
          \&= frac{1}{sqrt{left(frac{1+x^2}{1+x^2}right)^2-left(frac{2x}{1+x^2}right)^2}}cdotfrac{(1+x^2)(2) - (2x)(2x)}{(1+x^2)^2}
          \&= sqrt{frac{(1+x^2)^2}{(1-x^2)^2}}cdot frac{2(1-x^2)}{(1+x^2)^2}
          end{align*}
          (Skipping a bit of the algebra here). You would like to cancel the square and square root, and in the case of $(1+x^2)$, you can. But $1-x^2$ could be positive or negative, so $sqrt{(1-x^2)^2} = |1-x^2|$, not $1-x^2$. Therefore
          $$
          f'(x) = frac{2}{1+x^2}cdot frac{1-x^2}{|1-x^2|}
          $$
          As $xto 1$, the first factor tends to $2$, unambiguously. But the second factor is $pm 1$ depending on whether the numerator is positive or negative. That is,
          $$
          lim_{xto 1^-} f'(x) = 1qquadtext{and}qquadlim_{xto 1^+} f'(x) = -1
          $$
          It follows that $f$ cannot be differentiable at $1$.






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Recall that



            $$(arcsin u)'=frac1{sqrt{1-u^2}}$$



            and since the derivative tends to $infty$ as $uto 1^-$ the function is not differentiable at that point.



            That case is similar to $f(x)=sqrt x$ which has vertical tangent at $x=0$.






            share|cite|improve this answer











            $endgroup$













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              3 Answers
              3






              active

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              3 Answers
              3






              active

              oldest

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              active

              oldest

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              active

              oldest

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              1












              $begingroup$

              Note that $arcsin$ is differentiable in $(-1,1)$, but it is not differentiable at $pm1$. Since $frac{2times1}{1+1^2}=1$, it would be strange indeed if $f(x)=arcsinleft(frac{2x}{1+x^2}right)$ was differentiable at $1$.






              share|cite|improve this answer









              $endgroup$









              • 1




                $begingroup$
                What makes the function non differentiable at the extreme points.... Is there no neighbourhood to check verify slope of tangent at the extreme points?
                $endgroup$
                – GENESECT
                Aug 28 '18 at 14:29








              • 1




                $begingroup$
                The tangent lines to the graph of $arcsin$ at $pmleft(1,fracpi2right)$ are vertical lines. Therefore, $arcsin$ is not differentiable at $pm1$.
                $endgroup$
                – José Carlos Santos
                Aug 28 '18 at 14:31


















              1












              $begingroup$

              Note that $arcsin$ is differentiable in $(-1,1)$, but it is not differentiable at $pm1$. Since $frac{2times1}{1+1^2}=1$, it would be strange indeed if $f(x)=arcsinleft(frac{2x}{1+x^2}right)$ was differentiable at $1$.






              share|cite|improve this answer









              $endgroup$









              • 1




                $begingroup$
                What makes the function non differentiable at the extreme points.... Is there no neighbourhood to check verify slope of tangent at the extreme points?
                $endgroup$
                – GENESECT
                Aug 28 '18 at 14:29








              • 1




                $begingroup$
                The tangent lines to the graph of $arcsin$ at $pmleft(1,fracpi2right)$ are vertical lines. Therefore, $arcsin$ is not differentiable at $pm1$.
                $endgroup$
                – José Carlos Santos
                Aug 28 '18 at 14:31
















              1












              1








              1





              $begingroup$

              Note that $arcsin$ is differentiable in $(-1,1)$, but it is not differentiable at $pm1$. Since $frac{2times1}{1+1^2}=1$, it would be strange indeed if $f(x)=arcsinleft(frac{2x}{1+x^2}right)$ was differentiable at $1$.






              share|cite|improve this answer









              $endgroup$



              Note that $arcsin$ is differentiable in $(-1,1)$, but it is not differentiable at $pm1$. Since $frac{2times1}{1+1^2}=1$, it would be strange indeed if $f(x)=arcsinleft(frac{2x}{1+x^2}right)$ was differentiable at $1$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Aug 28 '18 at 14:27









              José Carlos SantosJosé Carlos Santos

              164k22132235




              164k22132235








              • 1




                $begingroup$
                What makes the function non differentiable at the extreme points.... Is there no neighbourhood to check verify slope of tangent at the extreme points?
                $endgroup$
                – GENESECT
                Aug 28 '18 at 14:29








              • 1




                $begingroup$
                The tangent lines to the graph of $arcsin$ at $pmleft(1,fracpi2right)$ are vertical lines. Therefore, $arcsin$ is not differentiable at $pm1$.
                $endgroup$
                – José Carlos Santos
                Aug 28 '18 at 14:31
















              • 1




                $begingroup$
                What makes the function non differentiable at the extreme points.... Is there no neighbourhood to check verify slope of tangent at the extreme points?
                $endgroup$
                – GENESECT
                Aug 28 '18 at 14:29








              • 1




                $begingroup$
                The tangent lines to the graph of $arcsin$ at $pmleft(1,fracpi2right)$ are vertical lines. Therefore, $arcsin$ is not differentiable at $pm1$.
                $endgroup$
                – José Carlos Santos
                Aug 28 '18 at 14:31










              1




              1




              $begingroup$
              What makes the function non differentiable at the extreme points.... Is there no neighbourhood to check verify slope of tangent at the extreme points?
              $endgroup$
              – GENESECT
              Aug 28 '18 at 14:29






              $begingroup$
              What makes the function non differentiable at the extreme points.... Is there no neighbourhood to check verify slope of tangent at the extreme points?
              $endgroup$
              – GENESECT
              Aug 28 '18 at 14:29






              1




              1




              $begingroup$
              The tangent lines to the graph of $arcsin$ at $pmleft(1,fracpi2right)$ are vertical lines. Therefore, $arcsin$ is not differentiable at $pm1$.
              $endgroup$
              – José Carlos Santos
              Aug 28 '18 at 14:31






              $begingroup$
              The tangent lines to the graph of $arcsin$ at $pmleft(1,fracpi2right)$ are vertical lines. Therefore, $arcsin$ is not differentiable at $pm1$.
              $endgroup$
              – José Carlos Santos
              Aug 28 '18 at 14:31













              1












              $begingroup$

              Notice that
              begin{align*}
              f'(x) &= frac{1}{sqrt{1-left(frac{2x}{1+x^2}right)^2}}frac{d}{dx}left(frac{2x}{1+x^2}right)
              \&= frac{1}{sqrt{left(frac{1+x^2}{1+x^2}right)^2-left(frac{2x}{1+x^2}right)^2}}cdotfrac{(1+x^2)(2) - (2x)(2x)}{(1+x^2)^2}
              \&= sqrt{frac{(1+x^2)^2}{(1-x^2)^2}}cdot frac{2(1-x^2)}{(1+x^2)^2}
              end{align*}
              (Skipping a bit of the algebra here). You would like to cancel the square and square root, and in the case of $(1+x^2)$, you can. But $1-x^2$ could be positive or negative, so $sqrt{(1-x^2)^2} = |1-x^2|$, not $1-x^2$. Therefore
              $$
              f'(x) = frac{2}{1+x^2}cdot frac{1-x^2}{|1-x^2|}
              $$
              As $xto 1$, the first factor tends to $2$, unambiguously. But the second factor is $pm 1$ depending on whether the numerator is positive or negative. That is,
              $$
              lim_{xto 1^-} f'(x) = 1qquadtext{and}qquadlim_{xto 1^+} f'(x) = -1
              $$
              It follows that $f$ cannot be differentiable at $1$.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                Notice that
                begin{align*}
                f'(x) &= frac{1}{sqrt{1-left(frac{2x}{1+x^2}right)^2}}frac{d}{dx}left(frac{2x}{1+x^2}right)
                \&= frac{1}{sqrt{left(frac{1+x^2}{1+x^2}right)^2-left(frac{2x}{1+x^2}right)^2}}cdotfrac{(1+x^2)(2) - (2x)(2x)}{(1+x^2)^2}
                \&= sqrt{frac{(1+x^2)^2}{(1-x^2)^2}}cdot frac{2(1-x^2)}{(1+x^2)^2}
                end{align*}
                (Skipping a bit of the algebra here). You would like to cancel the square and square root, and in the case of $(1+x^2)$, you can. But $1-x^2$ could be positive or negative, so $sqrt{(1-x^2)^2} = |1-x^2|$, not $1-x^2$. Therefore
                $$
                f'(x) = frac{2}{1+x^2}cdot frac{1-x^2}{|1-x^2|}
                $$
                As $xto 1$, the first factor tends to $2$, unambiguously. But the second factor is $pm 1$ depending on whether the numerator is positive or negative. That is,
                $$
                lim_{xto 1^-} f'(x) = 1qquadtext{and}qquadlim_{xto 1^+} f'(x) = -1
                $$
                It follows that $f$ cannot be differentiable at $1$.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  Notice that
                  begin{align*}
                  f'(x) &= frac{1}{sqrt{1-left(frac{2x}{1+x^2}right)^2}}frac{d}{dx}left(frac{2x}{1+x^2}right)
                  \&= frac{1}{sqrt{left(frac{1+x^2}{1+x^2}right)^2-left(frac{2x}{1+x^2}right)^2}}cdotfrac{(1+x^2)(2) - (2x)(2x)}{(1+x^2)^2}
                  \&= sqrt{frac{(1+x^2)^2}{(1-x^2)^2}}cdot frac{2(1-x^2)}{(1+x^2)^2}
                  end{align*}
                  (Skipping a bit of the algebra here). You would like to cancel the square and square root, and in the case of $(1+x^2)$, you can. But $1-x^2$ could be positive or negative, so $sqrt{(1-x^2)^2} = |1-x^2|$, not $1-x^2$. Therefore
                  $$
                  f'(x) = frac{2}{1+x^2}cdot frac{1-x^2}{|1-x^2|}
                  $$
                  As $xto 1$, the first factor tends to $2$, unambiguously. But the second factor is $pm 1$ depending on whether the numerator is positive or negative. That is,
                  $$
                  lim_{xto 1^-} f'(x) = 1qquadtext{and}qquadlim_{xto 1^+} f'(x) = -1
                  $$
                  It follows that $f$ cannot be differentiable at $1$.






                  share|cite|improve this answer









                  $endgroup$



                  Notice that
                  begin{align*}
                  f'(x) &= frac{1}{sqrt{1-left(frac{2x}{1+x^2}right)^2}}frac{d}{dx}left(frac{2x}{1+x^2}right)
                  \&= frac{1}{sqrt{left(frac{1+x^2}{1+x^2}right)^2-left(frac{2x}{1+x^2}right)^2}}cdotfrac{(1+x^2)(2) - (2x)(2x)}{(1+x^2)^2}
                  \&= sqrt{frac{(1+x^2)^2}{(1-x^2)^2}}cdot frac{2(1-x^2)}{(1+x^2)^2}
                  end{align*}
                  (Skipping a bit of the algebra here). You would like to cancel the square and square root, and in the case of $(1+x^2)$, you can. But $1-x^2$ could be positive or negative, so $sqrt{(1-x^2)^2} = |1-x^2|$, not $1-x^2$. Therefore
                  $$
                  f'(x) = frac{2}{1+x^2}cdot frac{1-x^2}{|1-x^2|}
                  $$
                  As $xto 1$, the first factor tends to $2$, unambiguously. But the second factor is $pm 1$ depending on whether the numerator is positive or negative. That is,
                  $$
                  lim_{xto 1^-} f'(x) = 1qquadtext{and}qquadlim_{xto 1^+} f'(x) = -1
                  $$
                  It follows that $f$ cannot be differentiable at $1$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Aug 28 '18 at 15:01









                  Matthew LeingangMatthew Leingang

                  16.8k12244




                  16.8k12244























                      0












                      $begingroup$

                      Recall that



                      $$(arcsin u)'=frac1{sqrt{1-u^2}}$$



                      and since the derivative tends to $infty$ as $uto 1^-$ the function is not differentiable at that point.



                      That case is similar to $f(x)=sqrt x$ which has vertical tangent at $x=0$.






                      share|cite|improve this answer











                      $endgroup$


















                        0












                        $begingroup$

                        Recall that



                        $$(arcsin u)'=frac1{sqrt{1-u^2}}$$



                        and since the derivative tends to $infty$ as $uto 1^-$ the function is not differentiable at that point.



                        That case is similar to $f(x)=sqrt x$ which has vertical tangent at $x=0$.






                        share|cite|improve this answer











                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          Recall that



                          $$(arcsin u)'=frac1{sqrt{1-u^2}}$$



                          and since the derivative tends to $infty$ as $uto 1^-$ the function is not differentiable at that point.



                          That case is similar to $f(x)=sqrt x$ which has vertical tangent at $x=0$.






                          share|cite|improve this answer











                          $endgroup$



                          Recall that



                          $$(arcsin u)'=frac1{sqrt{1-u^2}}$$



                          and since the derivative tends to $infty$ as $uto 1^-$ the function is not differentiable at that point.



                          That case is similar to $f(x)=sqrt x$ which has vertical tangent at $x=0$.







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Aug 28 '18 at 14:31

























                          answered Aug 28 '18 at 14:26









                          gimusigimusi

                          92.9k84494




                          92.9k84494






























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