Simple modification of complex expression












0












$begingroup$


I need to simplify
$$
{frac{1+itanalpha}{1-itanalpha}}
$$
I solved it:
$$
{frac{1+itanalpha}{1-itanalpha}}
= {frac{(1+itanalpha)(1+itanalpha)}{(1-itanalpha)(1+itanalpha)}}
= {frac{1-tan^2alpha+i(tanalpha+tanalpha)}{(1+itan^2alpha)(1+itanalpha)}}
= {frac{1-tan^2alpha+i(tanalpha+tanalpha)}{sec^2alpha}}
= {frac{1-tan^2alpha}{sec^2alpha}} + i{frac{2tanalpha}{sec^2alpha}}
$$
The answer in the book is
$$
cos2alpha+isin2alpha (alphane{frac{pi}{2}}+pi k)
$$
What's wrong here?






complex-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Are you sure it is ${frac{1+tanalpha}{1-itanalpha}}$ that has to be simplified and not ${frac{1+itanalpha}{1-itanalpha}}$?
    $endgroup$
    – mathcounterexamples.net
    Jan 19 at 18:16










  • $begingroup$
    Yes, sorry missed it. Updated the post.
    $endgroup$
    – user3132457
    Jan 19 at 18:17










  • $begingroup$
    Then it is quite easy... if you multiply the fraction by $cos alpha$.
    $endgroup$
    – mathcounterexamples.net
    Jan 19 at 18:17


















0












$begingroup$


I need to simplify
$$
{frac{1+itanalpha}{1-itanalpha}}
$$
I solved it:
$$
{frac{1+itanalpha}{1-itanalpha}}
= {frac{(1+itanalpha)(1+itanalpha)}{(1-itanalpha)(1+itanalpha)}}
= {frac{1-tan^2alpha+i(tanalpha+tanalpha)}{(1+itan^2alpha)(1+itanalpha)}}
= {frac{1-tan^2alpha+i(tanalpha+tanalpha)}{sec^2alpha}}
= {frac{1-tan^2alpha}{sec^2alpha}} + i{frac{2tanalpha}{sec^2alpha}}
$$
The answer in the book is
$$
cos2alpha+isin2alpha (alphane{frac{pi}{2}}+pi k)
$$
What's wrong here?






complex-numbers







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Are you sure it is ${frac{1+tanalpha}{1-itanalpha}}$ that has to be simplified and not ${frac{1+itanalpha}{1-itanalpha}}$?
    $endgroup$
    – mathcounterexamples.net
    Jan 19 at 18:16










  • $begingroup$
    Yes, sorry missed it. Updated the post.
    $endgroup$
    – user3132457
    Jan 19 at 18:17










  • $begingroup$
    Then it is quite easy... if you multiply the fraction by $cos alpha$.
    $endgroup$
    – mathcounterexamples.net
    Jan 19 at 18:17
















0












0








0





$begingroup$


I need to simplify
$$
{frac{1+itanalpha}{1-itanalpha}}
$$
I solved it:
$$
{frac{1+itanalpha}{1-itanalpha}}
= {frac{(1+itanalpha)(1+itanalpha)}{(1-itanalpha)(1+itanalpha)}}
= {frac{1-tan^2alpha+i(tanalpha+tanalpha)}{(1+itan^2alpha)(1+itanalpha)}}
= {frac{1-tan^2alpha+i(tanalpha+tanalpha)}{sec^2alpha}}
= {frac{1-tan^2alpha}{sec^2alpha}} + i{frac{2tanalpha}{sec^2alpha}}
$$
The answer in the book is
$$
cos2alpha+isin2alpha (alphane{frac{pi}{2}}+pi k)
$$
What's wrong here?






complex-numbers







share|cite|improve this question











$endgroup$




I need to simplify
$$
{frac{1+itanalpha}{1-itanalpha}}
$$
I solved it:
$$
{frac{1+itanalpha}{1-itanalpha}}
= {frac{(1+itanalpha)(1+itanalpha)}{(1-itanalpha)(1+itanalpha)}}
= {frac{1-tan^2alpha+i(tanalpha+tanalpha)}{(1+itan^2alpha)(1+itanalpha)}}
= {frac{1-tan^2alpha+i(tanalpha+tanalpha)}{sec^2alpha}}
= {frac{1-tan^2alpha}{sec^2alpha}} + i{frac{2tanalpha}{sec^2alpha}}
$$
The answer in the book is
$$
cos2alpha+isin2alpha (alphane{frac{pi}{2}}+pi k)
$$
What's wrong here?






complex-numbers




complex-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 19 at 18:17







user3132457

















asked Jan 19 at 18:14









user3132457user3132457

1598




1598








  • 1




    $begingroup$
    Are you sure it is ${frac{1+tanalpha}{1-itanalpha}}$ that has to be simplified and not ${frac{1+itanalpha}{1-itanalpha}}$?
    $endgroup$
    – mathcounterexamples.net
    Jan 19 at 18:16










  • $begingroup$
    Yes, sorry missed it. Updated the post.
    $endgroup$
    – user3132457
    Jan 19 at 18:17










  • $begingroup$
    Then it is quite easy... if you multiply the fraction by $cos alpha$.
    $endgroup$
    – mathcounterexamples.net
    Jan 19 at 18:17
















  • 1




    $begingroup$
    Are you sure it is ${frac{1+tanalpha}{1-itanalpha}}$ that has to be simplified and not ${frac{1+itanalpha}{1-itanalpha}}$?
    $endgroup$
    – mathcounterexamples.net
    Jan 19 at 18:16










  • $begingroup$
    Yes, sorry missed it. Updated the post.
    $endgroup$
    – user3132457
    Jan 19 at 18:17










  • $begingroup$
    Then it is quite easy... if you multiply the fraction by $cos alpha$.
    $endgroup$
    – mathcounterexamples.net
    Jan 19 at 18:17










1




1




$begingroup$
Are you sure it is ${frac{1+tanalpha}{1-itanalpha}}$ that has to be simplified and not ${frac{1+itanalpha}{1-itanalpha}}$?
$endgroup$
– mathcounterexamples.net
Jan 19 at 18:16




$begingroup$
Are you sure it is ${frac{1+tanalpha}{1-itanalpha}}$ that has to be simplified and not ${frac{1+itanalpha}{1-itanalpha}}$?
$endgroup$
– mathcounterexamples.net
Jan 19 at 18:16












$begingroup$
Yes, sorry missed it. Updated the post.
$endgroup$
– user3132457
Jan 19 at 18:17




$begingroup$
Yes, sorry missed it. Updated the post.
$endgroup$
– user3132457
Jan 19 at 18:17












$begingroup$
Then it is quite easy... if you multiply the fraction by $cos alpha$.
$endgroup$
– mathcounterexamples.net
Jan 19 at 18:17






$begingroup$
Then it is quite easy... if you multiply the fraction by $cos alpha$.
$endgroup$
– mathcounterexamples.net
Jan 19 at 18:17












4 Answers
4






active

oldest

votes


















2












$begingroup$

What you did is fine. Now$$frac{1-tan^2alpha}{sec^2alpha}=cos^2alpha-sin^2alpha=cos(2alpha)$$and$$2frac{tanalpha}{sec^2alpha}=2sinalphacosalpha=sin(2alpha).$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    You made some small typos which overall do not affect the final solution. I will write out a complete one to make everything more clear and convincible. Starting with the same as you we get
    begin{align}
    &color{red}{frac{1+itanalpha}{1-itanalpha}}=frac{1+itanalpha}{1-itanalpha}cdotfrac{1+itanalpha}{1+itanalpha}=frac{(1+itanalpha)^2}{1-i^2tan^2alpha}=frac{1-tan^2alpha+2itanalpha}{1+tan^2alpha}\
    &=frac{1-tan^2alpha}{1+tan^2alpha}+ifrac{2tanalpha}{1+tan^2alpha}=frac{1-frac{sin^2alpha}{cos^2alpha}}{1+frac{sin^2alpha}{cos^2alpha}}+ifrac{2frac{sinalpha}{cosalpha}}{1+frac{sin^2alpha}{cos^2alpha}}=frac{frac{cos^2alpha-sin^2alpha}{cos^2alpha}}{frac{cos^2alpha+sin^2alpha}{cos^2alpha}}+ifrac{2frac{sinalphacosalpha}{cos^2alpha}}{frac{cos^2alpha+sin^2alpha}{cos^2alpha}}\
    &=frac{cos^2alpha-sin^2alpha}{cos^2alpha+sin^2alpha}+ifrac{2sinalphacosalpha}{cos^2alpha+sin^2alpha}=frac{cos(2alpha)}1+ifrac{sin(2alpha)}1=color{red}{cos(2alpha)+isin(2alpha)}
    end{align}



    The restriction for $alpha$ is natural due the fact that for $alpha=fracpi2+kpi,~kinmathbb Z$ the tangent function has a pole of first order.






    share|cite|improve this answer











    $endgroup$





















      0












      $begingroup$

      Since $$frac{1-tan^2(x)}{sec^2(x)}=cos(2x)$$






      share|cite|improve this answer









      $endgroup$





















        0












        $begingroup$

        An easy and direct way is to multiply both the numerator and the denominator by $ cos alpha$. This can be done as long as $cosalphaneq 0$, i.e., $alphaneq frac{pi}{2} + pi k$.



        We obtain
        $$ frac{1+i tanalpha}{1-i tanalpha} = frac{cosalpha +i sinalpha}{cosalpha -i sinalpha} = frac{e^{ialpha}}{e^{-ialpha}} = e^{2ialpha} = cos(2alpha)+i sin(2alpha),.$$






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          How did you derive the first part of equation?
          $endgroup$
          – user3132457
          Jan 19 at 18:28






        • 2




          $begingroup$
          I simplify multiply the numerator and the denominator by $cos alpha$.
          $endgroup$
          – Fabian
          Jan 19 at 18:29











        Your Answer





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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        2












        $begingroup$

        What you did is fine. Now$$frac{1-tan^2alpha}{sec^2alpha}=cos^2alpha-sin^2alpha=cos(2alpha)$$and$$2frac{tanalpha}{sec^2alpha}=2sinalphacosalpha=sin(2alpha).$$






        share|cite|improve this answer









        $endgroup$


















          2












          $begingroup$

          What you did is fine. Now$$frac{1-tan^2alpha}{sec^2alpha}=cos^2alpha-sin^2alpha=cos(2alpha)$$and$$2frac{tanalpha}{sec^2alpha}=2sinalphacosalpha=sin(2alpha).$$






          share|cite|improve this answer









          $endgroup$
















            2












            2








            2





            $begingroup$

            What you did is fine. Now$$frac{1-tan^2alpha}{sec^2alpha}=cos^2alpha-sin^2alpha=cos(2alpha)$$and$$2frac{tanalpha}{sec^2alpha}=2sinalphacosalpha=sin(2alpha).$$






            share|cite|improve this answer









            $endgroup$



            What you did is fine. Now$$frac{1-tan^2alpha}{sec^2alpha}=cos^2alpha-sin^2alpha=cos(2alpha)$$and$$2frac{tanalpha}{sec^2alpha}=2sinalphacosalpha=sin(2alpha).$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 19 at 18:21









            José Carlos SantosJosé Carlos Santos

            162k22128232




            162k22128232























                1












                $begingroup$

                You made some small typos which overall do not affect the final solution. I will write out a complete one to make everything more clear and convincible. Starting with the same as you we get
                begin{align}
                &color{red}{frac{1+itanalpha}{1-itanalpha}}=frac{1+itanalpha}{1-itanalpha}cdotfrac{1+itanalpha}{1+itanalpha}=frac{(1+itanalpha)^2}{1-i^2tan^2alpha}=frac{1-tan^2alpha+2itanalpha}{1+tan^2alpha}\
                &=frac{1-tan^2alpha}{1+tan^2alpha}+ifrac{2tanalpha}{1+tan^2alpha}=frac{1-frac{sin^2alpha}{cos^2alpha}}{1+frac{sin^2alpha}{cos^2alpha}}+ifrac{2frac{sinalpha}{cosalpha}}{1+frac{sin^2alpha}{cos^2alpha}}=frac{frac{cos^2alpha-sin^2alpha}{cos^2alpha}}{frac{cos^2alpha+sin^2alpha}{cos^2alpha}}+ifrac{2frac{sinalphacosalpha}{cos^2alpha}}{frac{cos^2alpha+sin^2alpha}{cos^2alpha}}\
                &=frac{cos^2alpha-sin^2alpha}{cos^2alpha+sin^2alpha}+ifrac{2sinalphacosalpha}{cos^2alpha+sin^2alpha}=frac{cos(2alpha)}1+ifrac{sin(2alpha)}1=color{red}{cos(2alpha)+isin(2alpha)}
                end{align}



                The restriction for $alpha$ is natural due the fact that for $alpha=fracpi2+kpi,~kinmathbb Z$ the tangent function has a pole of first order.






                share|cite|improve this answer











                $endgroup$


















                  1












                  $begingroup$

                  You made some small typos which overall do not affect the final solution. I will write out a complete one to make everything more clear and convincible. Starting with the same as you we get
                  begin{align}
                  &color{red}{frac{1+itanalpha}{1-itanalpha}}=frac{1+itanalpha}{1-itanalpha}cdotfrac{1+itanalpha}{1+itanalpha}=frac{(1+itanalpha)^2}{1-i^2tan^2alpha}=frac{1-tan^2alpha+2itanalpha}{1+tan^2alpha}\
                  &=frac{1-tan^2alpha}{1+tan^2alpha}+ifrac{2tanalpha}{1+tan^2alpha}=frac{1-frac{sin^2alpha}{cos^2alpha}}{1+frac{sin^2alpha}{cos^2alpha}}+ifrac{2frac{sinalpha}{cosalpha}}{1+frac{sin^2alpha}{cos^2alpha}}=frac{frac{cos^2alpha-sin^2alpha}{cos^2alpha}}{frac{cos^2alpha+sin^2alpha}{cos^2alpha}}+ifrac{2frac{sinalphacosalpha}{cos^2alpha}}{frac{cos^2alpha+sin^2alpha}{cos^2alpha}}\
                  &=frac{cos^2alpha-sin^2alpha}{cos^2alpha+sin^2alpha}+ifrac{2sinalphacosalpha}{cos^2alpha+sin^2alpha}=frac{cos(2alpha)}1+ifrac{sin(2alpha)}1=color{red}{cos(2alpha)+isin(2alpha)}
                  end{align}



                  The restriction for $alpha$ is natural due the fact that for $alpha=fracpi2+kpi,~kinmathbb Z$ the tangent function has a pole of first order.






                  share|cite|improve this answer











                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    You made some small typos which overall do not affect the final solution. I will write out a complete one to make everything more clear and convincible. Starting with the same as you we get
                    begin{align}
                    &color{red}{frac{1+itanalpha}{1-itanalpha}}=frac{1+itanalpha}{1-itanalpha}cdotfrac{1+itanalpha}{1+itanalpha}=frac{(1+itanalpha)^2}{1-i^2tan^2alpha}=frac{1-tan^2alpha+2itanalpha}{1+tan^2alpha}\
                    &=frac{1-tan^2alpha}{1+tan^2alpha}+ifrac{2tanalpha}{1+tan^2alpha}=frac{1-frac{sin^2alpha}{cos^2alpha}}{1+frac{sin^2alpha}{cos^2alpha}}+ifrac{2frac{sinalpha}{cosalpha}}{1+frac{sin^2alpha}{cos^2alpha}}=frac{frac{cos^2alpha-sin^2alpha}{cos^2alpha}}{frac{cos^2alpha+sin^2alpha}{cos^2alpha}}+ifrac{2frac{sinalphacosalpha}{cos^2alpha}}{frac{cos^2alpha+sin^2alpha}{cos^2alpha}}\
                    &=frac{cos^2alpha-sin^2alpha}{cos^2alpha+sin^2alpha}+ifrac{2sinalphacosalpha}{cos^2alpha+sin^2alpha}=frac{cos(2alpha)}1+ifrac{sin(2alpha)}1=color{red}{cos(2alpha)+isin(2alpha)}
                    end{align}



                    The restriction for $alpha$ is natural due the fact that for $alpha=fracpi2+kpi,~kinmathbb Z$ the tangent function has a pole of first order.






                    share|cite|improve this answer











                    $endgroup$



                    You made some small typos which overall do not affect the final solution. I will write out a complete one to make everything more clear and convincible. Starting with the same as you we get
                    begin{align}
                    &color{red}{frac{1+itanalpha}{1-itanalpha}}=frac{1+itanalpha}{1-itanalpha}cdotfrac{1+itanalpha}{1+itanalpha}=frac{(1+itanalpha)^2}{1-i^2tan^2alpha}=frac{1-tan^2alpha+2itanalpha}{1+tan^2alpha}\
                    &=frac{1-tan^2alpha}{1+tan^2alpha}+ifrac{2tanalpha}{1+tan^2alpha}=frac{1-frac{sin^2alpha}{cos^2alpha}}{1+frac{sin^2alpha}{cos^2alpha}}+ifrac{2frac{sinalpha}{cosalpha}}{1+frac{sin^2alpha}{cos^2alpha}}=frac{frac{cos^2alpha-sin^2alpha}{cos^2alpha}}{frac{cos^2alpha+sin^2alpha}{cos^2alpha}}+ifrac{2frac{sinalphacosalpha}{cos^2alpha}}{frac{cos^2alpha+sin^2alpha}{cos^2alpha}}\
                    &=frac{cos^2alpha-sin^2alpha}{cos^2alpha+sin^2alpha}+ifrac{2sinalphacosalpha}{cos^2alpha+sin^2alpha}=frac{cos(2alpha)}1+ifrac{sin(2alpha)}1=color{red}{cos(2alpha)+isin(2alpha)}
                    end{align}



                    The restriction for $alpha$ is natural due the fact that for $alpha=fracpi2+kpi,~kinmathbb Z$ the tangent function has a pole of first order.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 20 at 0:47

























                    answered Jan 19 at 18:32









                    mrtaurhomrtaurho

                    5,46041237




                    5,46041237























                        0












                        $begingroup$

                        Since $$frac{1-tan^2(x)}{sec^2(x)}=cos(2x)$$






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Since $$frac{1-tan^2(x)}{sec^2(x)}=cos(2x)$$






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Since $$frac{1-tan^2(x)}{sec^2(x)}=cos(2x)$$






                            share|cite|improve this answer









                            $endgroup$



                            Since $$frac{1-tan^2(x)}{sec^2(x)}=cos(2x)$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Jan 19 at 18:18









                            Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                            75.6k42866




                            75.6k42866























                                0












                                $begingroup$

                                An easy and direct way is to multiply both the numerator and the denominator by $ cos alpha$. This can be done as long as $cosalphaneq 0$, i.e., $alphaneq frac{pi}{2} + pi k$.



                                We obtain
                                $$ frac{1+i tanalpha}{1-i tanalpha} = frac{cosalpha +i sinalpha}{cosalpha -i sinalpha} = frac{e^{ialpha}}{e^{-ialpha}} = e^{2ialpha} = cos(2alpha)+i sin(2alpha),.$$






                                share|cite|improve this answer









                                $endgroup$













                                • $begingroup$
                                  How did you derive the first part of equation?
                                  $endgroup$
                                  – user3132457
                                  Jan 19 at 18:28






                                • 2




                                  $begingroup$
                                  I simplify multiply the numerator and the denominator by $cos alpha$.
                                  $endgroup$
                                  – Fabian
                                  Jan 19 at 18:29
















                                0












                                $begingroup$

                                An easy and direct way is to multiply both the numerator and the denominator by $ cos alpha$. This can be done as long as $cosalphaneq 0$, i.e., $alphaneq frac{pi}{2} + pi k$.



                                We obtain
                                $$ frac{1+i tanalpha}{1-i tanalpha} = frac{cosalpha +i sinalpha}{cosalpha -i sinalpha} = frac{e^{ialpha}}{e^{-ialpha}} = e^{2ialpha} = cos(2alpha)+i sin(2alpha),.$$






                                share|cite|improve this answer









                                $endgroup$













                                • $begingroup$
                                  How did you derive the first part of equation?
                                  $endgroup$
                                  – user3132457
                                  Jan 19 at 18:28






                                • 2




                                  $begingroup$
                                  I simplify multiply the numerator and the denominator by $cos alpha$.
                                  $endgroup$
                                  – Fabian
                                  Jan 19 at 18:29














                                0












                                0








                                0





                                $begingroup$

                                An easy and direct way is to multiply both the numerator and the denominator by $ cos alpha$. This can be done as long as $cosalphaneq 0$, i.e., $alphaneq frac{pi}{2} + pi k$.



                                We obtain
                                $$ frac{1+i tanalpha}{1-i tanalpha} = frac{cosalpha +i sinalpha}{cosalpha -i sinalpha} = frac{e^{ialpha}}{e^{-ialpha}} = e^{2ialpha} = cos(2alpha)+i sin(2alpha),.$$






                                share|cite|improve this answer









                                $endgroup$



                                An easy and direct way is to multiply both the numerator and the denominator by $ cos alpha$. This can be done as long as $cosalphaneq 0$, i.e., $alphaneq frac{pi}{2} + pi k$.



                                We obtain
                                $$ frac{1+i tanalpha}{1-i tanalpha} = frac{cosalpha +i sinalpha}{cosalpha -i sinalpha} = frac{e^{ialpha}}{e^{-ialpha}} = e^{2ialpha} = cos(2alpha)+i sin(2alpha),.$$







                                share|cite|improve this answer












                                share|cite|improve this answer



                                share|cite|improve this answer










                                answered Jan 19 at 18:27









                                FabianFabian

                                19.8k3674




                                19.8k3674












                                • $begingroup$
                                  How did you derive the first part of equation?
                                  $endgroup$
                                  – user3132457
                                  Jan 19 at 18:28






                                • 2




                                  $begingroup$
                                  I simplify multiply the numerator and the denominator by $cos alpha$.
                                  $endgroup$
                                  – Fabian
                                  Jan 19 at 18:29


















                                • $begingroup$
                                  How did you derive the first part of equation?
                                  $endgroup$
                                  – user3132457
                                  Jan 19 at 18:28






                                • 2




                                  $begingroup$
                                  I simplify multiply the numerator and the denominator by $cos alpha$.
                                  $endgroup$
                                  – Fabian
                                  Jan 19 at 18:29
















                                $begingroup$
                                How did you derive the first part of equation?
                                $endgroup$
                                – user3132457
                                Jan 19 at 18:28




                                $begingroup$
                                How did you derive the first part of equation?
                                $endgroup$
                                – user3132457
                                Jan 19 at 18:28




                                2




                                2




                                $begingroup$
                                I simplify multiply the numerator and the denominator by $cos alpha$.
                                $endgroup$
                                – Fabian
                                Jan 19 at 18:29




                                $begingroup$
                                I simplify multiply the numerator and the denominator by $cos alpha$.
                                $endgroup$
                                – Fabian
                                Jan 19 at 18:29


















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                                Mario Kart Wii

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                                Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro...
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                                Partial Derivative Guidance.

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