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Differentiation

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dy/dx = (chain rule) 2nx(x2+1)n-1    
 
   When x=1, we have:  2n(2)n-1
   which equals: 4n(2)n-2 
   [since (2)n-1 = 2(2)n-2]
 
d2y/dx2=(product and chain rule) 2nx(2x)(n-1)(x2+1)n-2 +  2n(x2+1)n-1   
 
   When x=1, we have  2n(2(n-1)(2)n-2)+2n(2)n-1    (again: (2)n-1 = 2(2)n-1)
   which simplifies to 4n(n-1)(2)n-2+4n(2)n-2    
   leading to (2)n-2(4n(n-1) + 4n)   -{factoring out (2)n-2}
   4n(n-1) + 4n = 4n2 - 4n + 4n = 4n2
   SO, when x-1, d2y/dx2 = 4n2(2)n-2
 
 
 
Now, the ratio of dy/dx (when x=1) to d2y/dx2 (when x=1) is   
 
 
4n(2)n-2
___________
4n2(2)n-2
 
which equals (drum roll)  1:n