What am I doing wrong finding the derivative of $frac{3-x}{2}sqrt{1-2x-x^2}+2arcsin{frac{1+x}{sqrt{2}}}$?












1














$$y=frac{3-x}{2}sqrt{1-2x-x^2}+2arcsin{frac{1+x}{sqrt{2}}}$$
For convenience, let
$$A=frac{3-x}{2}sqrt{1-2x-x^2},$$
$$B=2arcsin{frac{1+x}{sqrt{2}}}.$$



$$y'=A'+B'$$
$$A'=(-frac{1}{2})(sqrt{1-2x-x^2})+(-2x)(frac{1}{2}cdot frac{1}{{sqrt{1-2x-x^2}}})(frac{3-x}{2})$$
$$A'=(-frac{1}{2})(sqrt{1-2x-x^2})+(-x)(frac{1}{{sqrt{1-2x-x^2}}})(frac{3-x}{2})$$
$$A'=(-frac{1-2x-x^2}{2sqrt{1-2x-x^2}})+(frac{x^2-3x}{{2sqrt{1-2x-x^2}}})$$
$$A'=(-frac{1-2x-x^2}{2sqrt{1-2x-x^2}})+(frac{x^2-3x}{{2sqrt{1-2x-x^2}}})$$
$$A'=frac{2x^2-x-1}{{2sqrt{1-2x-x^2}}}$$



$$B'=2bigg(arcsin{frac{1+x}{sqrt{2}}}bigg)'$$
$$B'=2 bigg( frac{1+x}{sqrt{2}} bigg)' bigg(arcsin{frac{1+x}{sqrt{2}}}bigg)'$$
$$B'=sqrt{2} bigg( frac{1}{sqrt{1- frac{1+2x+x^2}{2}} } bigg)$$
$$B'=sqrt{frac{4}{1-2x-x^2}}$$
$$B'=frac{4}{2sqrt{1-2x-x^2}}$$



$$y'=A'+B'=frac{2x^2-x+3}{2sqrt{1-2x-x^2}}$$



The answer in the book is
$$y'=frac{x^2}{sqrt{1-2x-x^2}}$$






calculus derivatives







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  • 1




    In the first line of your derivation of A': after the first summand, your second summand should be $$color{blue}{(-2 -2x)}(frac{1}{2}cdot frac{1}{{sqrt{1-2x-x^2}}})(frac{3-x}{2})$$
    – amWhy
    Nov 22 '18 at 22:24


















1














$$y=frac{3-x}{2}sqrt{1-2x-x^2}+2arcsin{frac{1+x}{sqrt{2}}}$$
For convenience, let
$$A=frac{3-x}{2}sqrt{1-2x-x^2},$$
$$B=2arcsin{frac{1+x}{sqrt{2}}}.$$



$$y'=A'+B'$$
$$A'=(-frac{1}{2})(sqrt{1-2x-x^2})+(-2x)(frac{1}{2}cdot frac{1}{{sqrt{1-2x-x^2}}})(frac{3-x}{2})$$
$$A'=(-frac{1}{2})(sqrt{1-2x-x^2})+(-x)(frac{1}{{sqrt{1-2x-x^2}}})(frac{3-x}{2})$$
$$A'=(-frac{1-2x-x^2}{2sqrt{1-2x-x^2}})+(frac{x^2-3x}{{2sqrt{1-2x-x^2}}})$$
$$A'=(-frac{1-2x-x^2}{2sqrt{1-2x-x^2}})+(frac{x^2-3x}{{2sqrt{1-2x-x^2}}})$$
$$A'=frac{2x^2-x-1}{{2sqrt{1-2x-x^2}}}$$



$$B'=2bigg(arcsin{frac{1+x}{sqrt{2}}}bigg)'$$
$$B'=2 bigg( frac{1+x}{sqrt{2}} bigg)' bigg(arcsin{frac{1+x}{sqrt{2}}}bigg)'$$
$$B'=sqrt{2} bigg( frac{1}{sqrt{1- frac{1+2x+x^2}{2}} } bigg)$$
$$B'=sqrt{frac{4}{1-2x-x^2}}$$
$$B'=frac{4}{2sqrt{1-2x-x^2}}$$



$$y'=A'+B'=frac{2x^2-x+3}{2sqrt{1-2x-x^2}}$$



The answer in the book is
$$y'=frac{x^2}{sqrt{1-2x-x^2}}$$






calculus derivatives







share|cite|improve this question


















  • 1




    In the first line of your derivation of A': after the first summand, your second summand should be $$color{blue}{(-2 -2x)}(frac{1}{2}cdot frac{1}{{sqrt{1-2x-x^2}}})(frac{3-x}{2})$$
    – amWhy
    Nov 22 '18 at 22:24
















1












1








1







$$y=frac{3-x}{2}sqrt{1-2x-x^2}+2arcsin{frac{1+x}{sqrt{2}}}$$
For convenience, let
$$A=frac{3-x}{2}sqrt{1-2x-x^2},$$
$$B=2arcsin{frac{1+x}{sqrt{2}}}.$$



$$y'=A'+B'$$
$$A'=(-frac{1}{2})(sqrt{1-2x-x^2})+(-2x)(frac{1}{2}cdot frac{1}{{sqrt{1-2x-x^2}}})(frac{3-x}{2})$$
$$A'=(-frac{1}{2})(sqrt{1-2x-x^2})+(-x)(frac{1}{{sqrt{1-2x-x^2}}})(frac{3-x}{2})$$
$$A'=(-frac{1-2x-x^2}{2sqrt{1-2x-x^2}})+(frac{x^2-3x}{{2sqrt{1-2x-x^2}}})$$
$$A'=(-frac{1-2x-x^2}{2sqrt{1-2x-x^2}})+(frac{x^2-3x}{{2sqrt{1-2x-x^2}}})$$
$$A'=frac{2x^2-x-1}{{2sqrt{1-2x-x^2}}}$$



$$B'=2bigg(arcsin{frac{1+x}{sqrt{2}}}bigg)'$$
$$B'=2 bigg( frac{1+x}{sqrt{2}} bigg)' bigg(arcsin{frac{1+x}{sqrt{2}}}bigg)'$$
$$B'=sqrt{2} bigg( frac{1}{sqrt{1- frac{1+2x+x^2}{2}} } bigg)$$
$$B'=sqrt{frac{4}{1-2x-x^2}}$$
$$B'=frac{4}{2sqrt{1-2x-x^2}}$$



$$y'=A'+B'=frac{2x^2-x+3}{2sqrt{1-2x-x^2}}$$



The answer in the book is
$$y'=frac{x^2}{sqrt{1-2x-x^2}}$$






calculus derivatives







share|cite|improve this question













$$y=frac{3-x}{2}sqrt{1-2x-x^2}+2arcsin{frac{1+x}{sqrt{2}}}$$
For convenience, let
$$A=frac{3-x}{2}sqrt{1-2x-x^2},$$
$$B=2arcsin{frac{1+x}{sqrt{2}}}.$$



$$y'=A'+B'$$
$$A'=(-frac{1}{2})(sqrt{1-2x-x^2})+(-2x)(frac{1}{2}cdot frac{1}{{sqrt{1-2x-x^2}}})(frac{3-x}{2})$$
$$A'=(-frac{1}{2})(sqrt{1-2x-x^2})+(-x)(frac{1}{{sqrt{1-2x-x^2}}})(frac{3-x}{2})$$
$$A'=(-frac{1-2x-x^2}{2sqrt{1-2x-x^2}})+(frac{x^2-3x}{{2sqrt{1-2x-x^2}}})$$
$$A'=(-frac{1-2x-x^2}{2sqrt{1-2x-x^2}})+(frac{x^2-3x}{{2sqrt{1-2x-x^2}}})$$
$$A'=frac{2x^2-x-1}{{2sqrt{1-2x-x^2}}}$$



$$B'=2bigg(arcsin{frac{1+x}{sqrt{2}}}bigg)'$$
$$B'=2 bigg( frac{1+x}{sqrt{2}} bigg)' bigg(arcsin{frac{1+x}{sqrt{2}}}bigg)'$$
$$B'=sqrt{2} bigg( frac{1}{sqrt{1- frac{1+2x+x^2}{2}} } bigg)$$
$$B'=sqrt{frac{4}{1-2x-x^2}}$$
$$B'=frac{4}{2sqrt{1-2x-x^2}}$$



$$y'=A'+B'=frac{2x^2-x+3}{2sqrt{1-2x-x^2}}$$



The answer in the book is
$$y'=frac{x^2}{sqrt{1-2x-x^2}}$$






calculus derivatives




calculus derivatives






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asked Nov 22 '18 at 22:17









fragileradiusfragileradius

297114




297114








  • 1




    In the first line of your derivation of A': after the first summand, your second summand should be $$color{blue}{(-2 -2x)}(frac{1}{2}cdot frac{1}{{sqrt{1-2x-x^2}}})(frac{3-x}{2})$$
    – amWhy
    Nov 22 '18 at 22:24
















  • 1




    In the first line of your derivation of A': after the first summand, your second summand should be $$color{blue}{(-2 -2x)}(frac{1}{2}cdot frac{1}{{sqrt{1-2x-x^2}}})(frac{3-x}{2})$$
    – amWhy
    Nov 22 '18 at 22:24










1




1




In the first line of your derivation of A': after the first summand, your second summand should be $$color{blue}{(-2 -2x)}(frac{1}{2}cdot frac{1}{{sqrt{1-2x-x^2}}})(frac{3-x}{2})$$
– amWhy
Nov 22 '18 at 22:24






In the first line of your derivation of A': after the first summand, your second summand should be $$color{blue}{(-2 -2x)}(frac{1}{2}cdot frac{1}{{sqrt{1-2x-x^2}}})(frac{3-x}{2})$$
– amWhy
Nov 22 '18 at 22:24












3 Answers
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You got A' wrong.



$ A = frac{3-x}{2}cdot(sqrt{1-2x-x^2}) = fcdot g$



where: $ f =frac{3-x}{2} $ , $ g = sqrt{1-2x-x^2} $



$A'= f'g + g'f$



$ f'g = (-frac{1}{2})(sqrt{1-2x-x^2})$ is correct, but the second part is $ g'f = (-2x -2)(frac{1}{2}cdot frac{1}{{sqrt{1-2x-x^2}}})(frac{3-x}{2})$ rather than
$(-2x)(frac{1}{2}cdot frac{1}{{sqrt{1-2x-x^2}}})(frac{3-x}{2})$



like you wrote.






share|cite|improve this answer





























    2














    In first $A'$ instead of $(-2x)$ you have $(-2-2x)$






    share|cite|improve this answer





























      0














      In the first line of your derivation of A': after the first summand, your second summand should be $$color{blue}{(-2 -2x)}left(frac{1}{2}cdot frac{1}{{sqrt{1-2x-x^2}}}right)left(frac{3-x}{2}right)$$



      So in fact, $$A'=left(-frac{1}{2}right)left(sqrt{1-2x-x^2}right)+ color{blue}{(-2 -2x)}left(frac{1}{2}cdot frac{1}{{sqrt{1-2x-x^2}}}right)left(frac{3-x}{2}right)$$






      share|cite|improve this answer























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        3 Answers
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        3 Answers
        3






        active

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        active

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        active

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        1














        You got A' wrong.



        $ A = frac{3-x}{2}cdot(sqrt{1-2x-x^2}) = fcdot g$



        where: $ f =frac{3-x}{2} $ , $ g = sqrt{1-2x-x^2} $



        $A'= f'g + g'f$



        $ f'g = (-frac{1}{2})(sqrt{1-2x-x^2})$ is correct, but the second part is $ g'f = (-2x -2)(frac{1}{2}cdot frac{1}{{sqrt{1-2x-x^2}}})(frac{3-x}{2})$ rather than
        $(-2x)(frac{1}{2}cdot frac{1}{{sqrt{1-2x-x^2}}})(frac{3-x}{2})$



        like you wrote.






        share|cite|improve this answer


























          1














          You got A' wrong.



          $ A = frac{3-x}{2}cdot(sqrt{1-2x-x^2}) = fcdot g$



          where: $ f =frac{3-x}{2} $ , $ g = sqrt{1-2x-x^2} $



          $A'= f'g + g'f$



          $ f'g = (-frac{1}{2})(sqrt{1-2x-x^2})$ is correct, but the second part is $ g'f = (-2x -2)(frac{1}{2}cdot frac{1}{{sqrt{1-2x-x^2}}})(frac{3-x}{2})$ rather than
          $(-2x)(frac{1}{2}cdot frac{1}{{sqrt{1-2x-x^2}}})(frac{3-x}{2})$



          like you wrote.






          share|cite|improve this answer
























            1












            1








            1






            You got A' wrong.



            $ A = frac{3-x}{2}cdot(sqrt{1-2x-x^2}) = fcdot g$



            where: $ f =frac{3-x}{2} $ , $ g = sqrt{1-2x-x^2} $



            $A'= f'g + g'f$



            $ f'g = (-frac{1}{2})(sqrt{1-2x-x^2})$ is correct, but the second part is $ g'f = (-2x -2)(frac{1}{2}cdot frac{1}{{sqrt{1-2x-x^2}}})(frac{3-x}{2})$ rather than
            $(-2x)(frac{1}{2}cdot frac{1}{{sqrt{1-2x-x^2}}})(frac{3-x}{2})$



            like you wrote.






            share|cite|improve this answer












            You got A' wrong.



            $ A = frac{3-x}{2}cdot(sqrt{1-2x-x^2}) = fcdot g$



            where: $ f =frac{3-x}{2} $ , $ g = sqrt{1-2x-x^2} $



            $A'= f'g + g'f$



            $ f'g = (-frac{1}{2})(sqrt{1-2x-x^2})$ is correct, but the second part is $ g'f = (-2x -2)(frac{1}{2}cdot frac{1}{{sqrt{1-2x-x^2}}})(frac{3-x}{2})$ rather than
            $(-2x)(frac{1}{2}cdot frac{1}{{sqrt{1-2x-x^2}}})(frac{3-x}{2})$



            like you wrote.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 22 '18 at 22:30









            Daphna KeidarDaphna Keidar

            1996




            1996























                2














                In first $A'$ instead of $(-2x)$ you have $(-2-2x)$






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                  2














                  In first $A'$ instead of $(-2x)$ you have $(-2-2x)$






                  share|cite|improve this answer
























                    2












                    2








                    2






                    In first $A'$ instead of $(-2x)$ you have $(-2-2x)$






                    share|cite|improve this answer












                    In first $A'$ instead of $(-2x)$ you have $(-2-2x)$







                    share|cite|improve this answer












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                    answered Nov 22 '18 at 22:25









                    asvasv

                    2841211




                    2841211























                        0














                        In the first line of your derivation of A': after the first summand, your second summand should be $$color{blue}{(-2 -2x)}left(frac{1}{2}cdot frac{1}{{sqrt{1-2x-x^2}}}right)left(frac{3-x}{2}right)$$



                        So in fact, $$A'=left(-frac{1}{2}right)left(sqrt{1-2x-x^2}right)+ color{blue}{(-2 -2x)}left(frac{1}{2}cdot frac{1}{{sqrt{1-2x-x^2}}}right)left(frac{3-x}{2}right)$$






                        share|cite|improve this answer




























                          0














                          In the first line of your derivation of A': after the first summand, your second summand should be $$color{blue}{(-2 -2x)}left(frac{1}{2}cdot frac{1}{{sqrt{1-2x-x^2}}}right)left(frac{3-x}{2}right)$$



                          So in fact, $$A'=left(-frac{1}{2}right)left(sqrt{1-2x-x^2}right)+ color{blue}{(-2 -2x)}left(frac{1}{2}cdot frac{1}{{sqrt{1-2x-x^2}}}right)left(frac{3-x}{2}right)$$






                          share|cite|improve this answer


























                            0












                            0








                            0






                            In the first line of your derivation of A': after the first summand, your second summand should be $$color{blue}{(-2 -2x)}left(frac{1}{2}cdot frac{1}{{sqrt{1-2x-x^2}}}right)left(frac{3-x}{2}right)$$



                            So in fact, $$A'=left(-frac{1}{2}right)left(sqrt{1-2x-x^2}right)+ color{blue}{(-2 -2x)}left(frac{1}{2}cdot frac{1}{{sqrt{1-2x-x^2}}}right)left(frac{3-x}{2}right)$$






                            share|cite|improve this answer














                            In the first line of your derivation of A': after the first summand, your second summand should be $$color{blue}{(-2 -2x)}left(frac{1}{2}cdot frac{1}{{sqrt{1-2x-x^2}}}right)left(frac{3-x}{2}right)$$



                            So in fact, $$A'=left(-frac{1}{2}right)left(sqrt{1-2x-x^2}right)+ color{blue}{(-2 -2x)}left(frac{1}{2}cdot frac{1}{{sqrt{1-2x-x^2}}}right)left(frac{3-x}{2}right)$$







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                            edited Nov 22 '18 at 22:47

























                            answered Nov 22 '18 at 22:28









                            amWhyamWhy

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                            192k28225439






























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