Derivative of $arctanbig (s-sqrt{1+s^{2}}big) $












-1














I am supposed to find the derivative of $f(s) = arctanbig(s-sqrt{1+s^{2}}big) $. My first step was this: $frac{1}{1+ (s-sqrt{1+s^{2}})^{2} } $. What am I supposed to do next? Thanks






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  • then you could try the chain rule and multiply by the derivative of $s-sqrt{1+s^2}$
    – Timothy Cho
    Nov 24 at 18:33










  • So I get $ frac{1}{1+ (s-sqrt{1+s^{2}})^{2} } * -1 ? $
    – Johny547
    Nov 24 at 18:46
















-1














I am supposed to find the derivative of $f(s) = arctanbig(s-sqrt{1+s^{2}}big) $. My first step was this: $frac{1}{1+ (s-sqrt{1+s^{2}})^{2} } $. What am I supposed to do next? Thanks






derivatives







share|cite|improve this question
























  • then you could try the chain rule and multiply by the derivative of $s-sqrt{1+s^2}$
    – Timothy Cho
    Nov 24 at 18:33










  • So I get $ frac{1}{1+ (s-sqrt{1+s^{2}})^{2} } * -1 ? $
    – Johny547
    Nov 24 at 18:46














-1












-1








-1







I am supposed to find the derivative of $f(s) = arctanbig(s-sqrt{1+s^{2}}big) $. My first step was this: $frac{1}{1+ (s-sqrt{1+s^{2}})^{2} } $. What am I supposed to do next? Thanks






derivatives







share|cite|improve this question















I am supposed to find the derivative of $f(s) = arctanbig(s-sqrt{1+s^{2}}big) $. My first step was this: $frac{1}{1+ (s-sqrt{1+s^{2}})^{2} } $. What am I supposed to do next? Thanks






derivatives




derivatives






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edited Nov 24 at 18:34









KM101

4,758421




4,758421










asked Nov 24 at 18:31









Johny547

1154




1154












  • then you could try the chain rule and multiply by the derivative of $s-sqrt{1+s^2}$
    – Timothy Cho
    Nov 24 at 18:33










  • So I get $ frac{1}{1+ (s-sqrt{1+s^{2}})^{2} } * -1 ? $
    – Johny547
    Nov 24 at 18:46


















  • then you could try the chain rule and multiply by the derivative of $s-sqrt{1+s^2}$
    – Timothy Cho
    Nov 24 at 18:33










  • So I get $ frac{1}{1+ (s-sqrt{1+s^{2}})^{2} } * -1 ? $
    – Johny547
    Nov 24 at 18:46
















then you could try the chain rule and multiply by the derivative of $s-sqrt{1+s^2}$
– Timothy Cho
Nov 24 at 18:33




then you could try the chain rule and multiply by the derivative of $s-sqrt{1+s^2}$
– Timothy Cho
Nov 24 at 18:33












So I get $ frac{1}{1+ (s-sqrt{1+s^{2}})^{2} } * -1 ? $
– Johny547
Nov 24 at 18:46




So I get $ frac{1}{1+ (s-sqrt{1+s^{2}})^{2} } * -1 ? $
– Johny547
Nov 24 at 18:46










2 Answers
2






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oldest

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0














May be this method is easier:



$$y=arctan(s-sqrt{1+s^2})$$



$$tan y=(s-sqrt{1+s^2})$$



Differentiating both sides we get:



$$(1+tan^2 y) dy=ds-frac{s}{sqrt{1+s^2}} ds$$



$$y'=frac{dy}{ds}=frac{1}{1+(s-sqrt{1+s^2})^2}.(1-frac{s}{sqrt{1+s^2}})$$






share|cite|improve this answer





























    0














    Use the Chain Rule.



    $$frac{d}{ds} arctan u = frac{1}{1+u^2}cdot color{blue}{frac{du}{ds}} tag{1}$$



    Let $u = s-sqrt{1+s^2}$ and solve for $color{blue}{frac{du}{ds}}$.



    $$color{blue}{frac{du}{ds} = frac{d}{ds}big(s-sqrt{1+s^2}big)}$$



    $$color{blue}{= 1-frac{d}{ds}sqrt{1+s^2}}$$



    Use the Chain Rule once more to find $color{purple}{frac{d}{ds}sqrt{1+s^2}}$.



    $$color{purple}{frac{d}{ds}sqrt{1+s^2} = frac{1}{2sqrt{1+s^2}}cdot2s}$$



    Returning to the blue expression, you get



    $$color{blue}{= 1-frac{2s}{2sqrt{1+s^2}} = 1-frac{s}{sqrt{1+s^2}}} tag{2}$$



    Returning to $(1)$, you get



    $$frac{d}{ds}arctanbig(s-sqrt{1+s^2}big) = frac{1}{1+big(s-sqrt{1+s^2}big)^2}cdotbigg(1-frac{s}{sqrt{1+s^2}}bigg)$$



    Which can be further simplified (not very easily and neatly).



    $$frac{1}{1+big(s-sqrt{1+s^2}big)^2}cdotbigg(1-frac{s}{sqrt{1+s^2}}bigg) = frac{1}{1+big(s-sqrt{1+s^2}big)^2}cdotbigg(frac{sqrt{1+s^2}-s}{sqrt{1+s^2}}bigg)$$



    $$= frac{(sqrt{1+s^2}-s}{big(1+big(s-sqrt{1+s^2}big)^2big)cdotbig(sqrt{1+s^2}big)}$$



    $$= frac{sqrt{1+s^2}-s}{big(1+s^2-2ssqrt{1+s^2}+1+s^2big)cdotbig(sqrt{1+s^2}big)}$$



    $$= frac{sqrt{1+s^2}-s}{big(2+2s^2-2ssqrt{1+s^2}big)cdotbig(sqrt{1+s^2}big)}$$



    $$= frac{sqrt{1+s^2}-s}{2sqrt{1+s^2}+2s^2sqrt{1+s^2}-2s(1+s^2)}$$



    $$= frac{sqrt{1+s^2}-s}{2big(sqrt{1+s^2}+s^2sqrt{1+s^2}-s(1+s^2)big)}$$



    $$= frac{sqrt{1+s^2}-s}{2big(sqrt{1+s^2}(1+s^2)-s(1+s^2)big)}$$



    $$= frac{sqrt{1+s^2}-s}{2big((1+s^2)big(sqrt{1+s^2}-sbig)big)} = frac{1}{2(s^2+1)} = frac{1}{2s^2+2}$$






    share|cite|improve this answer























    • How did you get those 2s? Also is this the final result?
      – Johny547
      Nov 24 at 18:55










    • I differentiated $s-sqrt{1+s^2}$ with respect to $s$ in order to find out $frac{du}{ds}$ to apply it to $(1)$, which gives the final answer (but it can be simplified).
      – KM101
      Nov 24 at 18:57












    • @Johny547 $$frac d{ds} sqrt{v}=frac 1{2sqrt{v}}frac{dv}{ds}$$ If $v=1+s^2$ ...
      – Andrei
      Nov 24 at 19:07













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    2 Answers
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    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes









    0














    May be this method is easier:



    $$y=arctan(s-sqrt{1+s^2})$$



    $$tan y=(s-sqrt{1+s^2})$$



    Differentiating both sides we get:



    $$(1+tan^2 y) dy=ds-frac{s}{sqrt{1+s^2}} ds$$



    $$y'=frac{dy}{ds}=frac{1}{1+(s-sqrt{1+s^2})^2}.(1-frac{s}{sqrt{1+s^2}})$$






    share|cite|improve this answer


























      0














      May be this method is easier:



      $$y=arctan(s-sqrt{1+s^2})$$



      $$tan y=(s-sqrt{1+s^2})$$



      Differentiating both sides we get:



      $$(1+tan^2 y) dy=ds-frac{s}{sqrt{1+s^2}} ds$$



      $$y'=frac{dy}{ds}=frac{1}{1+(s-sqrt{1+s^2})^2}.(1-frac{s}{sqrt{1+s^2}})$$






      share|cite|improve this answer
























        0












        0








        0






        May be this method is easier:



        $$y=arctan(s-sqrt{1+s^2})$$



        $$tan y=(s-sqrt{1+s^2})$$



        Differentiating both sides we get:



        $$(1+tan^2 y) dy=ds-frac{s}{sqrt{1+s^2}} ds$$



        $$y'=frac{dy}{ds}=frac{1}{1+(s-sqrt{1+s^2})^2}.(1-frac{s}{sqrt{1+s^2}})$$






        share|cite|improve this answer












        May be this method is easier:



        $$y=arctan(s-sqrt{1+s^2})$$



        $$tan y=(s-sqrt{1+s^2})$$



        Differentiating both sides we get:



        $$(1+tan^2 y) dy=ds-frac{s}{sqrt{1+s^2}} ds$$



        $$y'=frac{dy}{ds}=frac{1}{1+(s-sqrt{1+s^2})^2}.(1-frac{s}{sqrt{1+s^2}})$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 24 at 19:14









        sirous

        1,5991513




        1,5991513























            0














            Use the Chain Rule.



            $$frac{d}{ds} arctan u = frac{1}{1+u^2}cdot color{blue}{frac{du}{ds}} tag{1}$$



            Let $u = s-sqrt{1+s^2}$ and solve for $color{blue}{frac{du}{ds}}$.



            $$color{blue}{frac{du}{ds} = frac{d}{ds}big(s-sqrt{1+s^2}big)}$$



            $$color{blue}{= 1-frac{d}{ds}sqrt{1+s^2}}$$



            Use the Chain Rule once more to find $color{purple}{frac{d}{ds}sqrt{1+s^2}}$.



            $$color{purple}{frac{d}{ds}sqrt{1+s^2} = frac{1}{2sqrt{1+s^2}}cdot2s}$$



            Returning to the blue expression, you get



            $$color{blue}{= 1-frac{2s}{2sqrt{1+s^2}} = 1-frac{s}{sqrt{1+s^2}}} tag{2}$$



            Returning to $(1)$, you get



            $$frac{d}{ds}arctanbig(s-sqrt{1+s^2}big) = frac{1}{1+big(s-sqrt{1+s^2}big)^2}cdotbigg(1-frac{s}{sqrt{1+s^2}}bigg)$$



            Which can be further simplified (not very easily and neatly).



            $$frac{1}{1+big(s-sqrt{1+s^2}big)^2}cdotbigg(1-frac{s}{sqrt{1+s^2}}bigg) = frac{1}{1+big(s-sqrt{1+s^2}big)^2}cdotbigg(frac{sqrt{1+s^2}-s}{sqrt{1+s^2}}bigg)$$



            $$= frac{(sqrt{1+s^2}-s}{big(1+big(s-sqrt{1+s^2}big)^2big)cdotbig(sqrt{1+s^2}big)}$$



            $$= frac{sqrt{1+s^2}-s}{big(1+s^2-2ssqrt{1+s^2}+1+s^2big)cdotbig(sqrt{1+s^2}big)}$$



            $$= frac{sqrt{1+s^2}-s}{big(2+2s^2-2ssqrt{1+s^2}big)cdotbig(sqrt{1+s^2}big)}$$



            $$= frac{sqrt{1+s^2}-s}{2sqrt{1+s^2}+2s^2sqrt{1+s^2}-2s(1+s^2)}$$



            $$= frac{sqrt{1+s^2}-s}{2big(sqrt{1+s^2}+s^2sqrt{1+s^2}-s(1+s^2)big)}$$



            $$= frac{sqrt{1+s^2}-s}{2big(sqrt{1+s^2}(1+s^2)-s(1+s^2)big)}$$



            $$= frac{sqrt{1+s^2}-s}{2big((1+s^2)big(sqrt{1+s^2}-sbig)big)} = frac{1}{2(s^2+1)} = frac{1}{2s^2+2}$$






            share|cite|improve this answer























            • How did you get those 2s? Also is this the final result?
              – Johny547
              Nov 24 at 18:55










            • I differentiated $s-sqrt{1+s^2}$ with respect to $s$ in order to find out $frac{du}{ds}$ to apply it to $(1)$, which gives the final answer (but it can be simplified).
              – KM101
              Nov 24 at 18:57












            • @Johny547 $$frac d{ds} sqrt{v}=frac 1{2sqrt{v}}frac{dv}{ds}$$ If $v=1+s^2$ ...
              – Andrei
              Nov 24 at 19:07


















            0














            Use the Chain Rule.



            $$frac{d}{ds} arctan u = frac{1}{1+u^2}cdot color{blue}{frac{du}{ds}} tag{1}$$



            Let $u = s-sqrt{1+s^2}$ and solve for $color{blue}{frac{du}{ds}}$.



            $$color{blue}{frac{du}{ds} = frac{d}{ds}big(s-sqrt{1+s^2}big)}$$



            $$color{blue}{= 1-frac{d}{ds}sqrt{1+s^2}}$$



            Use the Chain Rule once more to find $color{purple}{frac{d}{ds}sqrt{1+s^2}}$.



            $$color{purple}{frac{d}{ds}sqrt{1+s^2} = frac{1}{2sqrt{1+s^2}}cdot2s}$$



            Returning to the blue expression, you get



            $$color{blue}{= 1-frac{2s}{2sqrt{1+s^2}} = 1-frac{s}{sqrt{1+s^2}}} tag{2}$$



            Returning to $(1)$, you get



            $$frac{d}{ds}arctanbig(s-sqrt{1+s^2}big) = frac{1}{1+big(s-sqrt{1+s^2}big)^2}cdotbigg(1-frac{s}{sqrt{1+s^2}}bigg)$$



            Which can be further simplified (not very easily and neatly).



            $$frac{1}{1+big(s-sqrt{1+s^2}big)^2}cdotbigg(1-frac{s}{sqrt{1+s^2}}bigg) = frac{1}{1+big(s-sqrt{1+s^2}big)^2}cdotbigg(frac{sqrt{1+s^2}-s}{sqrt{1+s^2}}bigg)$$



            $$= frac{(sqrt{1+s^2}-s}{big(1+big(s-sqrt{1+s^2}big)^2big)cdotbig(sqrt{1+s^2}big)}$$



            $$= frac{sqrt{1+s^2}-s}{big(1+s^2-2ssqrt{1+s^2}+1+s^2big)cdotbig(sqrt{1+s^2}big)}$$



            $$= frac{sqrt{1+s^2}-s}{big(2+2s^2-2ssqrt{1+s^2}big)cdotbig(sqrt{1+s^2}big)}$$



            $$= frac{sqrt{1+s^2}-s}{2sqrt{1+s^2}+2s^2sqrt{1+s^2}-2s(1+s^2)}$$



            $$= frac{sqrt{1+s^2}-s}{2big(sqrt{1+s^2}+s^2sqrt{1+s^2}-s(1+s^2)big)}$$



            $$= frac{sqrt{1+s^2}-s}{2big(sqrt{1+s^2}(1+s^2)-s(1+s^2)big)}$$



            $$= frac{sqrt{1+s^2}-s}{2big((1+s^2)big(sqrt{1+s^2}-sbig)big)} = frac{1}{2(s^2+1)} = frac{1}{2s^2+2}$$






            share|cite|improve this answer























            • How did you get those 2s? Also is this the final result?
              – Johny547
              Nov 24 at 18:55










            • I differentiated $s-sqrt{1+s^2}$ with respect to $s$ in order to find out $frac{du}{ds}$ to apply it to $(1)$, which gives the final answer (but it can be simplified).
              – KM101
              Nov 24 at 18:57












            • @Johny547 $$frac d{ds} sqrt{v}=frac 1{2sqrt{v}}frac{dv}{ds}$$ If $v=1+s^2$ ...
              – Andrei
              Nov 24 at 19:07
















            0












            0








            0






            Use the Chain Rule.



            $$frac{d}{ds} arctan u = frac{1}{1+u^2}cdot color{blue}{frac{du}{ds}} tag{1}$$



            Let $u = s-sqrt{1+s^2}$ and solve for $color{blue}{frac{du}{ds}}$.



            $$color{blue}{frac{du}{ds} = frac{d}{ds}big(s-sqrt{1+s^2}big)}$$



            $$color{blue}{= 1-frac{d}{ds}sqrt{1+s^2}}$$



            Use the Chain Rule once more to find $color{purple}{frac{d}{ds}sqrt{1+s^2}}$.



            $$color{purple}{frac{d}{ds}sqrt{1+s^2} = frac{1}{2sqrt{1+s^2}}cdot2s}$$



            Returning to the blue expression, you get



            $$color{blue}{= 1-frac{2s}{2sqrt{1+s^2}} = 1-frac{s}{sqrt{1+s^2}}} tag{2}$$



            Returning to $(1)$, you get



            $$frac{d}{ds}arctanbig(s-sqrt{1+s^2}big) = frac{1}{1+big(s-sqrt{1+s^2}big)^2}cdotbigg(1-frac{s}{sqrt{1+s^2}}bigg)$$



            Which can be further simplified (not very easily and neatly).



            $$frac{1}{1+big(s-sqrt{1+s^2}big)^2}cdotbigg(1-frac{s}{sqrt{1+s^2}}bigg) = frac{1}{1+big(s-sqrt{1+s^2}big)^2}cdotbigg(frac{sqrt{1+s^2}-s}{sqrt{1+s^2}}bigg)$$



            $$= frac{(sqrt{1+s^2}-s}{big(1+big(s-sqrt{1+s^2}big)^2big)cdotbig(sqrt{1+s^2}big)}$$



            $$= frac{sqrt{1+s^2}-s}{big(1+s^2-2ssqrt{1+s^2}+1+s^2big)cdotbig(sqrt{1+s^2}big)}$$



            $$= frac{sqrt{1+s^2}-s}{big(2+2s^2-2ssqrt{1+s^2}big)cdotbig(sqrt{1+s^2}big)}$$



            $$= frac{sqrt{1+s^2}-s}{2sqrt{1+s^2}+2s^2sqrt{1+s^2}-2s(1+s^2)}$$



            $$= frac{sqrt{1+s^2}-s}{2big(sqrt{1+s^2}+s^2sqrt{1+s^2}-s(1+s^2)big)}$$



            $$= frac{sqrt{1+s^2}-s}{2big(sqrt{1+s^2}(1+s^2)-s(1+s^2)big)}$$



            $$= frac{sqrt{1+s^2}-s}{2big((1+s^2)big(sqrt{1+s^2}-sbig)big)} = frac{1}{2(s^2+1)} = frac{1}{2s^2+2}$$






            share|cite|improve this answer














            Use the Chain Rule.



            $$frac{d}{ds} arctan u = frac{1}{1+u^2}cdot color{blue}{frac{du}{ds}} tag{1}$$



            Let $u = s-sqrt{1+s^2}$ and solve for $color{blue}{frac{du}{ds}}$.



            $$color{blue}{frac{du}{ds} = frac{d}{ds}big(s-sqrt{1+s^2}big)}$$



            $$color{blue}{= 1-frac{d}{ds}sqrt{1+s^2}}$$



            Use the Chain Rule once more to find $color{purple}{frac{d}{ds}sqrt{1+s^2}}$.



            $$color{purple}{frac{d}{ds}sqrt{1+s^2} = frac{1}{2sqrt{1+s^2}}cdot2s}$$



            Returning to the blue expression, you get



            $$color{blue}{= 1-frac{2s}{2sqrt{1+s^2}} = 1-frac{s}{sqrt{1+s^2}}} tag{2}$$



            Returning to $(1)$, you get



            $$frac{d}{ds}arctanbig(s-sqrt{1+s^2}big) = frac{1}{1+big(s-sqrt{1+s^2}big)^2}cdotbigg(1-frac{s}{sqrt{1+s^2}}bigg)$$



            Which can be further simplified (not very easily and neatly).



            $$frac{1}{1+big(s-sqrt{1+s^2}big)^2}cdotbigg(1-frac{s}{sqrt{1+s^2}}bigg) = frac{1}{1+big(s-sqrt{1+s^2}big)^2}cdotbigg(frac{sqrt{1+s^2}-s}{sqrt{1+s^2}}bigg)$$



            $$= frac{(sqrt{1+s^2}-s}{big(1+big(s-sqrt{1+s^2}big)^2big)cdotbig(sqrt{1+s^2}big)}$$



            $$= frac{sqrt{1+s^2}-s}{big(1+s^2-2ssqrt{1+s^2}+1+s^2big)cdotbig(sqrt{1+s^2}big)}$$



            $$= frac{sqrt{1+s^2}-s}{big(2+2s^2-2ssqrt{1+s^2}big)cdotbig(sqrt{1+s^2}big)}$$



            $$= frac{sqrt{1+s^2}-s}{2sqrt{1+s^2}+2s^2sqrt{1+s^2}-2s(1+s^2)}$$



            $$= frac{sqrt{1+s^2}-s}{2big(sqrt{1+s^2}+s^2sqrt{1+s^2}-s(1+s^2)big)}$$



            $$= frac{sqrt{1+s^2}-s}{2big(sqrt{1+s^2}(1+s^2)-s(1+s^2)big)}$$



            $$= frac{sqrt{1+s^2}-s}{2big((1+s^2)big(sqrt{1+s^2}-sbig)big)} = frac{1}{2(s^2+1)} = frac{1}{2s^2+2}$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Nov 24 at 19:54

























            answered Nov 24 at 18:49









            KM101

            4,758421




            4,758421












            • How did you get those 2s? Also is this the final result?
              – Johny547
              Nov 24 at 18:55










            • I differentiated $s-sqrt{1+s^2}$ with respect to $s$ in order to find out $frac{du}{ds}$ to apply it to $(1)$, which gives the final answer (but it can be simplified).
              – KM101
              Nov 24 at 18:57












            • @Johny547 $$frac d{ds} sqrt{v}=frac 1{2sqrt{v}}frac{dv}{ds}$$ If $v=1+s^2$ ...
              – Andrei
              Nov 24 at 19:07




















            • How did you get those 2s? Also is this the final result?
              – Johny547
              Nov 24 at 18:55










            • I differentiated $s-sqrt{1+s^2}$ with respect to $s$ in order to find out $frac{du}{ds}$ to apply it to $(1)$, which gives the final answer (but it can be simplified).
              – KM101
              Nov 24 at 18:57












            • @Johny547 $$frac d{ds} sqrt{v}=frac 1{2sqrt{v}}frac{dv}{ds}$$ If $v=1+s^2$ ...
              – Andrei
              Nov 24 at 19:07


















            How did you get those 2s? Also is this the final result?
            – Johny547
            Nov 24 at 18:55




            How did you get those 2s? Also is this the final result?
            – Johny547
            Nov 24 at 18:55












            I differentiated $s-sqrt{1+s^2}$ with respect to $s$ in order to find out $frac{du}{ds}$ to apply it to $(1)$, which gives the final answer (but it can be simplified).
            – KM101
            Nov 24 at 18:57






            I differentiated $s-sqrt{1+s^2}$ with respect to $s$ in order to find out $frac{du}{ds}$ to apply it to $(1)$, which gives the final answer (but it can be simplified).
            – KM101
            Nov 24 at 18:57














            @Johny547 $$frac d{ds} sqrt{v}=frac 1{2sqrt{v}}frac{dv}{ds}$$ If $v=1+s^2$ ...
            – Andrei
            Nov 24 at 19:07






            @Johny547 $$frac d{ds} sqrt{v}=frac 1{2sqrt{v}}frac{dv}{ds}$$ If $v=1+s^2$ ...
            – Andrei
            Nov 24 at 19:07




















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