Differentiate $sin^{-1}left(frac {sin x + cos x}{sqrt{2}}right)$ with respect to $x$












1












$begingroup$



Differentiate $$sin^{-1}left(frac {sin x + cos x}{sqrt{2}}right)$$ with respect to $x$.




I started like this: Consider $$frac {sin x + cos x}{sqrt{2}}$$, substitute $cos x$ as $sin left(frac {pi}{2} - xright)$, and proceed with the simplification. Finally I am getting it as $cos left(x - frac {pi}{4}right)$. After this I could not proceed. Any help would be appreciated. Thanks in advance!






calculus derivatives







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$endgroup$












  • $begingroup$
    @user222031 I also tried like that but I ended up with $frac {cos x - sinx}{sqrt (1 - sin 2x)}$. How to proceed after this?
    $endgroup$
    – Sahana
    May 5 '15 at 13:28










  • $begingroup$
    $sin x cos pi/4 + cos x sin pi/4$
    $endgroup$
    – Mann
    May 5 '15 at 13:29












  • $begingroup$
    @mann Where to use this? Sorry I don't get it
    $endgroup$
    – Sahana
    May 5 '15 at 13:30










  • $begingroup$
    Look the answers below, you'd see it.
    $endgroup$
    – Mann
    May 5 '15 at 13:31
















1












$begingroup$



Differentiate $$sin^{-1}left(frac {sin x + cos x}{sqrt{2}}right)$$ with respect to $x$.




I started like this: Consider $$frac {sin x + cos x}{sqrt{2}}$$, substitute $cos x$ as $sin left(frac {pi}{2} - xright)$, and proceed with the simplification. Finally I am getting it as $cos left(x - frac {pi}{4}right)$. After this I could not proceed. Any help would be appreciated. Thanks in advance!






calculus derivatives







share|cite|improve this question











$endgroup$












  • $begingroup$
    @user222031 I also tried like that but I ended up with $frac {cos x - sinx}{sqrt (1 - sin 2x)}$. How to proceed after this?
    $endgroup$
    – Sahana
    May 5 '15 at 13:28










  • $begingroup$
    $sin x cos pi/4 + cos x sin pi/4$
    $endgroup$
    – Mann
    May 5 '15 at 13:29












  • $begingroup$
    @mann Where to use this? Sorry I don't get it
    $endgroup$
    – Sahana
    May 5 '15 at 13:30










  • $begingroup$
    Look the answers below, you'd see it.
    $endgroup$
    – Mann
    May 5 '15 at 13:31














1












1








1


1



$begingroup$



Differentiate $$sin^{-1}left(frac {sin x + cos x}{sqrt{2}}right)$$ with respect to $x$.




I started like this: Consider $$frac {sin x + cos x}{sqrt{2}}$$, substitute $cos x$ as $sin left(frac {pi}{2} - xright)$, and proceed with the simplification. Finally I am getting it as $cos left(x - frac {pi}{4}right)$. After this I could not proceed. Any help would be appreciated. Thanks in advance!






calculus derivatives







share|cite|improve this question











$endgroup$





Differentiate $$sin^{-1}left(frac {sin x + cos x}{sqrt{2}}right)$$ with respect to $x$.




I started like this: Consider $$frac {sin x + cos x}{sqrt{2}}$$, substitute $cos x$ as $sin left(frac {pi}{2} - xright)$, and proceed with the simplification. Finally I am getting it as $cos left(x - frac {pi}{4}right)$. After this I could not proceed. Any help would be appreciated. Thanks in advance!






calculus derivatives




calculus derivatives






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited May 5 '15 at 13:30









Travis

63.7k769151




63.7k769151










asked May 5 '15 at 13:21









SahanaSahana

2416




2416












  • $begingroup$
    @user222031 I also tried like that but I ended up with $frac {cos x - sinx}{sqrt (1 - sin 2x)}$. How to proceed after this?
    $endgroup$
    – Sahana
    May 5 '15 at 13:28










  • $begingroup$
    $sin x cos pi/4 + cos x sin pi/4$
    $endgroup$
    – Mann
    May 5 '15 at 13:29












  • $begingroup$
    @mann Where to use this? Sorry I don't get it
    $endgroup$
    – Sahana
    May 5 '15 at 13:30










  • $begingroup$
    Look the answers below, you'd see it.
    $endgroup$
    – Mann
    May 5 '15 at 13:31


















  • $begingroup$
    @user222031 I also tried like that but I ended up with $frac {cos x - sinx}{sqrt (1 - sin 2x)}$. How to proceed after this?
    $endgroup$
    – Sahana
    May 5 '15 at 13:28










  • $begingroup$
    $sin x cos pi/4 + cos x sin pi/4$
    $endgroup$
    – Mann
    May 5 '15 at 13:29












  • $begingroup$
    @mann Where to use this? Sorry I don't get it
    $endgroup$
    – Sahana
    May 5 '15 at 13:30










  • $begingroup$
    Look the answers below, you'd see it.
    $endgroup$
    – Mann
    May 5 '15 at 13:31
















$begingroup$
@user222031 I also tried like that but I ended up with $frac {cos x - sinx}{sqrt (1 - sin 2x)}$. How to proceed after this?
$endgroup$
– Sahana
May 5 '15 at 13:28




$begingroup$
@user222031 I also tried like that but I ended up with $frac {cos x - sinx}{sqrt (1 - sin 2x)}$. How to proceed after this?
$endgroup$
– Sahana
May 5 '15 at 13:28












$begingroup$
$sin x cos pi/4 + cos x sin pi/4$
$endgroup$
– Mann
May 5 '15 at 13:29






$begingroup$
$sin x cos pi/4 + cos x sin pi/4$
$endgroup$
– Mann
May 5 '15 at 13:29














$begingroup$
@mann Where to use this? Sorry I don't get it
$endgroup$
– Sahana
May 5 '15 at 13:30




$begingroup$
@mann Where to use this? Sorry I don't get it
$endgroup$
– Sahana
May 5 '15 at 13:30












$begingroup$
Look the answers below, you'd see it.
$endgroup$
– Mann
May 5 '15 at 13:31




$begingroup$
Look the answers below, you'd see it.
$endgroup$
– Mann
May 5 '15 at 13:31










3 Answers
3






active

oldest

votes


















5












$begingroup$

Hint The angle sum rule for $sin$ is
$$sin(x + y) = sin x cos y + cos x sin y.$$




Additional hint In particular, if we take $y = frac{pi}{4}$ and rearrange, we get $$sin x + cos x = sqrt{2} sinleft(x + frac{pi}{4}right).$$







share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    It's better to rewrite
    $$
    frac{1}{sqrt{2}}(sin x+cos x)=sin(x+pi/4)
    $$
    and then use the chain rule:
    $$
    f(x)=arcsinsinBigl(x+frac{pi}{4}Bigr)
    $$
    so
    $$
    f'(x)=frac{1}{sqrt{1-sin^2(x+pi/4)}}cosBigl(x+frac{pi}{4}Bigr)
    =dots
    $$
    (Beware of the square root!)




    $f'(x)=dfrac{cos(x+pi/4)}{|cos(x+pi/4)|}$

    so the derivative is $1$ where $cos(x+pi/4)>0$ and $-1$ where $cos(x+pi/4)<0$; the function is not differentiable where $cos(x+pi/4)=0$.







    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Then $sin^{-1} Sin (x + frac {pi}{4})$ is simply $x + frac {pi}{4}$ and if we differentiate it we will get 1. Am I right?
      $endgroup$
      – Sahana
      May 5 '15 at 13:33










    • $begingroup$
      @Sahana The equality $arcsinsinalpha=alpha$ is generally false. For instance, $arcsinsinpi=arcsin 0=0nepi$.
      $endgroup$
      – egreg
      May 5 '15 at 13:34












    • $begingroup$
      BEWARE OF THE SQUARE ROOT.
      $endgroup$
      – Mann
      May 5 '15 at 13:35










    • $begingroup$
      If I simplify @egreg expression also the final ans is coming as $1$. So I think in this case $sin^{-1}(sin x) = x$. is it ok?
      $endgroup$
      – Sahana
      May 5 '15 at 13:38










    • $begingroup$
      Still no. Greg final expression is not always 1.
      $endgroup$
      – Mann
      May 5 '15 at 13:40



















    0












    $begingroup$

    An alternative approach is to use Implicit Differentiation:



    begin{equation}
    y = arcsinleft(frac{sin(x) + cos(x)}{sqrt{2}} right) rightarrow sin(y) = frac{sin(x) + cos(x)}{sqrt{2}}
    end{equation}



    Now differentiate with respect to '$x$':



    begin{align}
    frac{d}{dx}left[sin(y) right] &= frac{d}{dx}left[frac{sin(x) + cos(x)}{sqrt{2}} right] \
    cos(y)frac{dy}{dx} &= frac{cos(x) - sin(x)}{sqrt{2}} \
    frac{dy}{dx} &= frac{cos(x) - sin(x)}{sqrt{2}cos(y)}
    end{align}



    Thus:



    begin{equation}
    frac{dy}{dx} = frac{d}{dx}left[arcsinleft(frac{sin(x) + cos(x)}{sqrt{2}} right) right] = frac{cos(x) - sin(x)}{sqrt{2}cosleft(arcsinleft(frac{sin(x) + cos(x)}{sqrt{2}}right) right)}
    end{equation}



    Here this method is unnecessarily complicated in comparison to those already presented. It is however good to know if an identity is either unknown.






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5












      $begingroup$

      Hint The angle sum rule for $sin$ is
      $$sin(x + y) = sin x cos y + cos x sin y.$$




      Additional hint In particular, if we take $y = frac{pi}{4}$ and rearrange, we get $$sin x + cos x = sqrt{2} sinleft(x + frac{pi}{4}right).$$







      share|cite|improve this answer









      $endgroup$


















        5












        $begingroup$

        Hint The angle sum rule for $sin$ is
        $$sin(x + y) = sin x cos y + cos x sin y.$$




        Additional hint In particular, if we take $y = frac{pi}{4}$ and rearrange, we get $$sin x + cos x = sqrt{2} sinleft(x + frac{pi}{4}right).$$







        share|cite|improve this answer









        $endgroup$
















          5












          5








          5





          $begingroup$

          Hint The angle sum rule for $sin$ is
          $$sin(x + y) = sin x cos y + cos x sin y.$$




          Additional hint In particular, if we take $y = frac{pi}{4}$ and rearrange, we get $$sin x + cos x = sqrt{2} sinleft(x + frac{pi}{4}right).$$







          share|cite|improve this answer









          $endgroup$



          Hint The angle sum rule for $sin$ is
          $$sin(x + y) = sin x cos y + cos x sin y.$$




          Additional hint In particular, if we take $y = frac{pi}{4}$ and rearrange, we get $$sin x + cos x = sqrt{2} sinleft(x + frac{pi}{4}right).$$








          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered May 5 '15 at 13:27









          TravisTravis

          63.7k769151




          63.7k769151























              2












              $begingroup$

              It's better to rewrite
              $$
              frac{1}{sqrt{2}}(sin x+cos x)=sin(x+pi/4)
              $$
              and then use the chain rule:
              $$
              f(x)=arcsinsinBigl(x+frac{pi}{4}Bigr)
              $$
              so
              $$
              f'(x)=frac{1}{sqrt{1-sin^2(x+pi/4)}}cosBigl(x+frac{pi}{4}Bigr)
              =dots
              $$
              (Beware of the square root!)




              $f'(x)=dfrac{cos(x+pi/4)}{|cos(x+pi/4)|}$

              so the derivative is $1$ where $cos(x+pi/4)>0$ and $-1$ where $cos(x+pi/4)<0$; the function is not differentiable where $cos(x+pi/4)=0$.







              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Then $sin^{-1} Sin (x + frac {pi}{4})$ is simply $x + frac {pi}{4}$ and if we differentiate it we will get 1. Am I right?
                $endgroup$
                – Sahana
                May 5 '15 at 13:33










              • $begingroup$
                @Sahana The equality $arcsinsinalpha=alpha$ is generally false. For instance, $arcsinsinpi=arcsin 0=0nepi$.
                $endgroup$
                – egreg
                May 5 '15 at 13:34












              • $begingroup$
                BEWARE OF THE SQUARE ROOT.
                $endgroup$
                – Mann
                May 5 '15 at 13:35










              • $begingroup$
                If I simplify @egreg expression also the final ans is coming as $1$. So I think in this case $sin^{-1}(sin x) = x$. is it ok?
                $endgroup$
                – Sahana
                May 5 '15 at 13:38










              • $begingroup$
                Still no. Greg final expression is not always 1.
                $endgroup$
                – Mann
                May 5 '15 at 13:40
















              2












              $begingroup$

              It's better to rewrite
              $$
              frac{1}{sqrt{2}}(sin x+cos x)=sin(x+pi/4)
              $$
              and then use the chain rule:
              $$
              f(x)=arcsinsinBigl(x+frac{pi}{4}Bigr)
              $$
              so
              $$
              f'(x)=frac{1}{sqrt{1-sin^2(x+pi/4)}}cosBigl(x+frac{pi}{4}Bigr)
              =dots
              $$
              (Beware of the square root!)




              $f'(x)=dfrac{cos(x+pi/4)}{|cos(x+pi/4)|}$

              so the derivative is $1$ where $cos(x+pi/4)>0$ and $-1$ where $cos(x+pi/4)<0$; the function is not differentiable where $cos(x+pi/4)=0$.







              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Then $sin^{-1} Sin (x + frac {pi}{4})$ is simply $x + frac {pi}{4}$ and if we differentiate it we will get 1. Am I right?
                $endgroup$
                – Sahana
                May 5 '15 at 13:33










              • $begingroup$
                @Sahana The equality $arcsinsinalpha=alpha$ is generally false. For instance, $arcsinsinpi=arcsin 0=0nepi$.
                $endgroup$
                – egreg
                May 5 '15 at 13:34












              • $begingroup$
                BEWARE OF THE SQUARE ROOT.
                $endgroup$
                – Mann
                May 5 '15 at 13:35










              • $begingroup$
                If I simplify @egreg expression also the final ans is coming as $1$. So I think in this case $sin^{-1}(sin x) = x$. is it ok?
                $endgroup$
                – Sahana
                May 5 '15 at 13:38










              • $begingroup$
                Still no. Greg final expression is not always 1.
                $endgroup$
                – Mann
                May 5 '15 at 13:40














              2












              2








              2





              $begingroup$

              It's better to rewrite
              $$
              frac{1}{sqrt{2}}(sin x+cos x)=sin(x+pi/4)
              $$
              and then use the chain rule:
              $$
              f(x)=arcsinsinBigl(x+frac{pi}{4}Bigr)
              $$
              so
              $$
              f'(x)=frac{1}{sqrt{1-sin^2(x+pi/4)}}cosBigl(x+frac{pi}{4}Bigr)
              =dots
              $$
              (Beware of the square root!)




              $f'(x)=dfrac{cos(x+pi/4)}{|cos(x+pi/4)|}$

              so the derivative is $1$ where $cos(x+pi/4)>0$ and $-1$ where $cos(x+pi/4)<0$; the function is not differentiable where $cos(x+pi/4)=0$.







              share|cite|improve this answer











              $endgroup$



              It's better to rewrite
              $$
              frac{1}{sqrt{2}}(sin x+cos x)=sin(x+pi/4)
              $$
              and then use the chain rule:
              $$
              f(x)=arcsinsinBigl(x+frac{pi}{4}Bigr)
              $$
              so
              $$
              f'(x)=frac{1}{sqrt{1-sin^2(x+pi/4)}}cosBigl(x+frac{pi}{4}Bigr)
              =dots
              $$
              (Beware of the square root!)




              $f'(x)=dfrac{cos(x+pi/4)}{|cos(x+pi/4)|}$

              so the derivative is $1$ where $cos(x+pi/4)>0$ and $-1$ where $cos(x+pi/4)<0$; the function is not differentiable where $cos(x+pi/4)=0$.








              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited May 5 '15 at 13:45

























              answered May 5 '15 at 13:28









              egregegreg

              184k1486206




              184k1486206












              • $begingroup$
                Then $sin^{-1} Sin (x + frac {pi}{4})$ is simply $x + frac {pi}{4}$ and if we differentiate it we will get 1. Am I right?
                $endgroup$
                – Sahana
                May 5 '15 at 13:33










              • $begingroup$
                @Sahana The equality $arcsinsinalpha=alpha$ is generally false. For instance, $arcsinsinpi=arcsin 0=0nepi$.
                $endgroup$
                – egreg
                May 5 '15 at 13:34












              • $begingroup$
                BEWARE OF THE SQUARE ROOT.
                $endgroup$
                – Mann
                May 5 '15 at 13:35










              • $begingroup$
                If I simplify @egreg expression also the final ans is coming as $1$. So I think in this case $sin^{-1}(sin x) = x$. is it ok?
                $endgroup$
                – Sahana
                May 5 '15 at 13:38










              • $begingroup$
                Still no. Greg final expression is not always 1.
                $endgroup$
                – Mann
                May 5 '15 at 13:40


















              • $begingroup$
                Then $sin^{-1} Sin (x + frac {pi}{4})$ is simply $x + frac {pi}{4}$ and if we differentiate it we will get 1. Am I right?
                $endgroup$
                – Sahana
                May 5 '15 at 13:33










              • $begingroup$
                @Sahana The equality $arcsinsinalpha=alpha$ is generally false. For instance, $arcsinsinpi=arcsin 0=0nepi$.
                $endgroup$
                – egreg
                May 5 '15 at 13:34












              • $begingroup$
                BEWARE OF THE SQUARE ROOT.
                $endgroup$
                – Mann
                May 5 '15 at 13:35










              • $begingroup$
                If I simplify @egreg expression also the final ans is coming as $1$. So I think in this case $sin^{-1}(sin x) = x$. is it ok?
                $endgroup$
                – Sahana
                May 5 '15 at 13:38










              • $begingroup$
                Still no. Greg final expression is not always 1.
                $endgroup$
                – Mann
                May 5 '15 at 13:40
















              $begingroup$
              Then $sin^{-1} Sin (x + frac {pi}{4})$ is simply $x + frac {pi}{4}$ and if we differentiate it we will get 1. Am I right?
              $endgroup$
              – Sahana
              May 5 '15 at 13:33




              $begingroup$
              Then $sin^{-1} Sin (x + frac {pi}{4})$ is simply $x + frac {pi}{4}$ and if we differentiate it we will get 1. Am I right?
              $endgroup$
              – Sahana
              May 5 '15 at 13:33












              $begingroup$
              @Sahana The equality $arcsinsinalpha=alpha$ is generally false. For instance, $arcsinsinpi=arcsin 0=0nepi$.
              $endgroup$
              – egreg
              May 5 '15 at 13:34






              $begingroup$
              @Sahana The equality $arcsinsinalpha=alpha$ is generally false. For instance, $arcsinsinpi=arcsin 0=0nepi$.
              $endgroup$
              – egreg
              May 5 '15 at 13:34














              $begingroup$
              BEWARE OF THE SQUARE ROOT.
              $endgroup$
              – Mann
              May 5 '15 at 13:35




              $begingroup$
              BEWARE OF THE SQUARE ROOT.
              $endgroup$
              – Mann
              May 5 '15 at 13:35












              $begingroup$
              If I simplify @egreg expression also the final ans is coming as $1$. So I think in this case $sin^{-1}(sin x) = x$. is it ok?
              $endgroup$
              – Sahana
              May 5 '15 at 13:38




              $begingroup$
              If I simplify @egreg expression also the final ans is coming as $1$. So I think in this case $sin^{-1}(sin x) = x$. is it ok?
              $endgroup$
              – Sahana
              May 5 '15 at 13:38












              $begingroup$
              Still no. Greg final expression is not always 1.
              $endgroup$
              – Mann
              May 5 '15 at 13:40




              $begingroup$
              Still no. Greg final expression is not always 1.
              $endgroup$
              – Mann
              May 5 '15 at 13:40











              0












              $begingroup$

              An alternative approach is to use Implicit Differentiation:



              begin{equation}
              y = arcsinleft(frac{sin(x) + cos(x)}{sqrt{2}} right) rightarrow sin(y) = frac{sin(x) + cos(x)}{sqrt{2}}
              end{equation}



              Now differentiate with respect to '$x$':



              begin{align}
              frac{d}{dx}left[sin(y) right] &= frac{d}{dx}left[frac{sin(x) + cos(x)}{sqrt{2}} right] \
              cos(y)frac{dy}{dx} &= frac{cos(x) - sin(x)}{sqrt{2}} \
              frac{dy}{dx} &= frac{cos(x) - sin(x)}{sqrt{2}cos(y)}
              end{align}



              Thus:



              begin{equation}
              frac{dy}{dx} = frac{d}{dx}left[arcsinleft(frac{sin(x) + cos(x)}{sqrt{2}} right) right] = frac{cos(x) - sin(x)}{sqrt{2}cosleft(arcsinleft(frac{sin(x) + cos(x)}{sqrt{2}}right) right)}
              end{equation}



              Here this method is unnecessarily complicated in comparison to those already presented. It is however good to know if an identity is either unknown.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                An alternative approach is to use Implicit Differentiation:



                begin{equation}
                y = arcsinleft(frac{sin(x) + cos(x)}{sqrt{2}} right) rightarrow sin(y) = frac{sin(x) + cos(x)}{sqrt{2}}
                end{equation}



                Now differentiate with respect to '$x$':



                begin{align}
                frac{d}{dx}left[sin(y) right] &= frac{d}{dx}left[frac{sin(x) + cos(x)}{sqrt{2}} right] \
                cos(y)frac{dy}{dx} &= frac{cos(x) - sin(x)}{sqrt{2}} \
                frac{dy}{dx} &= frac{cos(x) - sin(x)}{sqrt{2}cos(y)}
                end{align}



                Thus:



                begin{equation}
                frac{dy}{dx} = frac{d}{dx}left[arcsinleft(frac{sin(x) + cos(x)}{sqrt{2}} right) right] = frac{cos(x) - sin(x)}{sqrt{2}cosleft(arcsinleft(frac{sin(x) + cos(x)}{sqrt{2}}right) right)}
                end{equation}



                Here this method is unnecessarily complicated in comparison to those already presented. It is however good to know if an identity is either unknown.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  An alternative approach is to use Implicit Differentiation:



                  begin{equation}
                  y = arcsinleft(frac{sin(x) + cos(x)}{sqrt{2}} right) rightarrow sin(y) = frac{sin(x) + cos(x)}{sqrt{2}}
                  end{equation}



                  Now differentiate with respect to '$x$':



                  begin{align}
                  frac{d}{dx}left[sin(y) right] &= frac{d}{dx}left[frac{sin(x) + cos(x)}{sqrt{2}} right] \
                  cos(y)frac{dy}{dx} &= frac{cos(x) - sin(x)}{sqrt{2}} \
                  frac{dy}{dx} &= frac{cos(x) - sin(x)}{sqrt{2}cos(y)}
                  end{align}



                  Thus:



                  begin{equation}
                  frac{dy}{dx} = frac{d}{dx}left[arcsinleft(frac{sin(x) + cos(x)}{sqrt{2}} right) right] = frac{cos(x) - sin(x)}{sqrt{2}cosleft(arcsinleft(frac{sin(x) + cos(x)}{sqrt{2}}right) right)}
                  end{equation}



                  Here this method is unnecessarily complicated in comparison to those already presented. It is however good to know if an identity is either unknown.






                  share|cite|improve this answer









                  $endgroup$



                  An alternative approach is to use Implicit Differentiation:



                  begin{equation}
                  y = arcsinleft(frac{sin(x) + cos(x)}{sqrt{2}} right) rightarrow sin(y) = frac{sin(x) + cos(x)}{sqrt{2}}
                  end{equation}



                  Now differentiate with respect to '$x$':



                  begin{align}
                  frac{d}{dx}left[sin(y) right] &= frac{d}{dx}left[frac{sin(x) + cos(x)}{sqrt{2}} right] \
                  cos(y)frac{dy}{dx} &= frac{cos(x) - sin(x)}{sqrt{2}} \
                  frac{dy}{dx} &= frac{cos(x) - sin(x)}{sqrt{2}cos(y)}
                  end{align}



                  Thus:



                  begin{equation}
                  frac{dy}{dx} = frac{d}{dx}left[arcsinleft(frac{sin(x) + cos(x)}{sqrt{2}} right) right] = frac{cos(x) - sin(x)}{sqrt{2}cosleft(arcsinleft(frac{sin(x) + cos(x)}{sqrt{2}}right) right)}
                  end{equation}



                  Here this method is unnecessarily complicated in comparison to those already presented. It is however good to know if an identity is either unknown.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 6 at 12:19









                  DavidGDavidG

                  1




                  1






























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