Determine the equation of the normal to the tangent curve

First:  Find the y-value when x = 3:  y  =  (3·3)/(32 - 1)  =  9/8   --->   (3, 9/8)

 

Second:  Find the first derivative of y:

     y'  =  [ (x2 - 1)·(3)  -  (3x)·(2x) ] [ (x2 - 1)2 ]  =  [ -3x2 - 3 ] / [ (x2 - 1)2 ]

 

Third:  Find the value of the first derivative when x = 3:  y'  =  [ -3(3)2 - 3 ] / [ ((3)2 - 1)2 ]  =  -15/32

 

Fourth:  Find the slope of the normal (perpendicular) at that point:  slope  =  32/15

 

Fifth:  Find the equation of the normal by using the point-slope form:  point = (3, 9/8)  m  =  32/15

      y - 9/8  =  (32/15)(x - 3)