How do I find an equation of the tangent line to the graph?












1














Find an equation of the tangent line to the graph of $arctan(x+y)=y^2+frac{pi}{4}$ at $(1, 0)$.



This is what I have so far:



$1 + frac{dy}{dx}=2ycdotsec^2(y^2+frac{pi}{4})cdotfrac{dy}{dx}$



How do I go on from here?






calculus







share|cite|improve this question





























    1














    Find an equation of the tangent line to the graph of $arctan(x+y)=y^2+frac{pi}{4}$ at $(1, 0)$.



    This is what I have so far:



    $1 + frac{dy}{dx}=2ycdotsec^2(y^2+frac{pi}{4})cdotfrac{dy}{dx}$



    How do I go on from here?






    calculus







    share|cite|improve this question



























      1












      1








      1







      Find an equation of the tangent line to the graph of $arctan(x+y)=y^2+frac{pi}{4}$ at $(1, 0)$.



      This is what I have so far:



      $1 + frac{dy}{dx}=2ycdotsec^2(y^2+frac{pi}{4})cdotfrac{dy}{dx}$



      How do I go on from here?






      calculus







      share|cite|improve this question















      Find an equation of the tangent line to the graph of $arctan(x+y)=y^2+frac{pi}{4}$ at $(1, 0)$.



      This is what I have so far:



      $1 + frac{dy}{dx}=2ycdotsec^2(y^2+frac{pi}{4})cdotfrac{dy}{dx}$



      How do I go on from here?






      calculus




      calculus






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 26 at 17:38









      ArsenBerk

      7,55131338




      7,55131338










      asked Nov 26 at 17:31









      user597553

      213




      213






















          3 Answers
          3






          active

          oldest

          votes


















          1














          So from what you have, get $frac{dy}{dx}$ by itself. If you do that you should have



          $$frac{dy}{dx}=frac{1}{2ycdot sec^2{left(y^2+frac{pi}{4}right)-1}}$$



          Now, there is no need to try and simplify since you are only interested in a tangent line. Replace all $x$'s with $1$ and all $y$'s with $0$. Then the slope of the tangent line is



          $$frac{1}{2cdot0cdot sec^2{left(0^2+frac{pi}{4}right)-1}}=frac{1}{-1}=-1$$



          Thus the equation of the tangent line is



          $$y-0=-1(x-1)rightarrow y=1-x$$






          share|cite|improve this answer

















          • 1




            Thank you for helping!
            – user597553
            Nov 26 at 18:03



















          0














          Tangent line is described by first derivative. Basically, you should find y', then find y' at the point and you will have the slope of the tangent line.



          Thus, y' (1,0) is slope of line. Use linear equation for the line, $y_0=kx_0+b$. Subscript 0 is about the point (1,0); k is the slope. Then extrapolate as you know k and b.



          Can you (as you know the slope of line at the point) find equation? If you need furhter help, leave a comment below.






          share|cite|improve this answer





















          • Thank you for helping!
            – user597553
            Nov 26 at 18:03



















          0














          $ ( x=1,y=0) $ to be plugged in into the relation you already got by differentiation, to evaluate slope at tangent point.. then and there as $ ( y^{'}=-1)$. Luckily $y=0$ no need to find actual graph by integration.



          Equation of tangent (point-slope form) is



          $$ dfrac{y-0}{x-1}=-1 rightarrow x+y=1. $$






          share|cite|improve this answer





















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            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1














            So from what you have, get $frac{dy}{dx}$ by itself. If you do that you should have



            $$frac{dy}{dx}=frac{1}{2ycdot sec^2{left(y^2+frac{pi}{4}right)-1}}$$



            Now, there is no need to try and simplify since you are only interested in a tangent line. Replace all $x$'s with $1$ and all $y$'s with $0$. Then the slope of the tangent line is



            $$frac{1}{2cdot0cdot sec^2{left(0^2+frac{pi}{4}right)-1}}=frac{1}{-1}=-1$$



            Thus the equation of the tangent line is



            $$y-0=-1(x-1)rightarrow y=1-x$$






            share|cite|improve this answer

















            • 1




              Thank you for helping!
              – user597553
              Nov 26 at 18:03
















            1














            So from what you have, get $frac{dy}{dx}$ by itself. If you do that you should have



            $$frac{dy}{dx}=frac{1}{2ycdot sec^2{left(y^2+frac{pi}{4}right)-1}}$$



            Now, there is no need to try and simplify since you are only interested in a tangent line. Replace all $x$'s with $1$ and all $y$'s with $0$. Then the slope of the tangent line is



            $$frac{1}{2cdot0cdot sec^2{left(0^2+frac{pi}{4}right)-1}}=frac{1}{-1}=-1$$



            Thus the equation of the tangent line is



            $$y-0=-1(x-1)rightarrow y=1-x$$






            share|cite|improve this answer

















            • 1




              Thank you for helping!
              – user597553
              Nov 26 at 18:03














            1












            1








            1






            So from what you have, get $frac{dy}{dx}$ by itself. If you do that you should have



            $$frac{dy}{dx}=frac{1}{2ycdot sec^2{left(y^2+frac{pi}{4}right)-1}}$$



            Now, there is no need to try and simplify since you are only interested in a tangent line. Replace all $x$'s with $1$ and all $y$'s with $0$. Then the slope of the tangent line is



            $$frac{1}{2cdot0cdot sec^2{left(0^2+frac{pi}{4}right)-1}}=frac{1}{-1}=-1$$



            Thus the equation of the tangent line is



            $$y-0=-1(x-1)rightarrow y=1-x$$






            share|cite|improve this answer












            So from what you have, get $frac{dy}{dx}$ by itself. If you do that you should have



            $$frac{dy}{dx}=frac{1}{2ycdot sec^2{left(y^2+frac{pi}{4}right)-1}}$$



            Now, there is no need to try and simplify since you are only interested in a tangent line. Replace all $x$'s with $1$ and all $y$'s with $0$. Then the slope of the tangent line is



            $$frac{1}{2cdot0cdot sec^2{left(0^2+frac{pi}{4}right)-1}}=frac{1}{-1}=-1$$



            Thus the equation of the tangent line is



            $$y-0=-1(x-1)rightarrow y=1-x$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 26 at 18:00









            Eleven-Eleven

            5,39072659




            5,39072659








            • 1




              Thank you for helping!
              – user597553
              Nov 26 at 18:03














            • 1




              Thank you for helping!
              – user597553
              Nov 26 at 18:03








            1




            1




            Thank you for helping!
            – user597553
            Nov 26 at 18:03




            Thank you for helping!
            – user597553
            Nov 26 at 18:03











            0














            Tangent line is described by first derivative. Basically, you should find y', then find y' at the point and you will have the slope of the tangent line.



            Thus, y' (1,0) is slope of line. Use linear equation for the line, $y_0=kx_0+b$. Subscript 0 is about the point (1,0); k is the slope. Then extrapolate as you know k and b.



            Can you (as you know the slope of line at the point) find equation? If you need furhter help, leave a comment below.






            share|cite|improve this answer





















            • Thank you for helping!
              – user597553
              Nov 26 at 18:03
















            0














            Tangent line is described by first derivative. Basically, you should find y', then find y' at the point and you will have the slope of the tangent line.



            Thus, y' (1,0) is slope of line. Use linear equation for the line, $y_0=kx_0+b$. Subscript 0 is about the point (1,0); k is the slope. Then extrapolate as you know k and b.



            Can you (as you know the slope of line at the point) find equation? If you need furhter help, leave a comment below.






            share|cite|improve this answer





















            • Thank you for helping!
              – user597553
              Nov 26 at 18:03














            0












            0








            0






            Tangent line is described by first derivative. Basically, you should find y', then find y' at the point and you will have the slope of the tangent line.



            Thus, y' (1,0) is slope of line. Use linear equation for the line, $y_0=kx_0+b$. Subscript 0 is about the point (1,0); k is the slope. Then extrapolate as you know k and b.



            Can you (as you know the slope of line at the point) find equation? If you need furhter help, leave a comment below.






            share|cite|improve this answer












            Tangent line is described by first derivative. Basically, you should find y', then find y' at the point and you will have the slope of the tangent line.



            Thus, y' (1,0) is slope of line. Use linear equation for the line, $y_0=kx_0+b$. Subscript 0 is about the point (1,0); k is the slope. Then extrapolate as you know k and b.



            Can you (as you know the slope of line at the point) find equation? If you need furhter help, leave a comment below.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 26 at 17:47









            Kelly Shepphard

            2298




            2298












            • Thank you for helping!
              – user597553
              Nov 26 at 18:03


















            • Thank you for helping!
              – user597553
              Nov 26 at 18:03
















            Thank you for helping!
            – user597553
            Nov 26 at 18:03




            Thank you for helping!
            – user597553
            Nov 26 at 18:03











            0














            $ ( x=1,y=0) $ to be plugged in into the relation you already got by differentiation, to evaluate slope at tangent point.. then and there as $ ( y^{'}=-1)$. Luckily $y=0$ no need to find actual graph by integration.



            Equation of tangent (point-slope form) is



            $$ dfrac{y-0}{x-1}=-1 rightarrow x+y=1. $$






            share|cite|improve this answer


























              0














              $ ( x=1,y=0) $ to be plugged in into the relation you already got by differentiation, to evaluate slope at tangent point.. then and there as $ ( y^{'}=-1)$. Luckily $y=0$ no need to find actual graph by integration.



              Equation of tangent (point-slope form) is



              $$ dfrac{y-0}{x-1}=-1 rightarrow x+y=1. $$






              share|cite|improve this answer
























                0












                0








                0






                $ ( x=1,y=0) $ to be plugged in into the relation you already got by differentiation, to evaluate slope at tangent point.. then and there as $ ( y^{'}=-1)$. Luckily $y=0$ no need to find actual graph by integration.



                Equation of tangent (point-slope form) is



                $$ dfrac{y-0}{x-1}=-1 rightarrow x+y=1. $$






                share|cite|improve this answer












                $ ( x=1,y=0) $ to be plugged in into the relation you already got by differentiation, to evaluate slope at tangent point.. then and there as $ ( y^{'}=-1)$. Luckily $y=0$ no need to find actual graph by integration.



                Equation of tangent (point-slope form) is



                $$ dfrac{y-0}{x-1}=-1 rightarrow x+y=1. $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 26 at 19:29









                Narasimham

                20.6k52158




                20.6k52158






























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