Differentiation

Question

Given that y= (x²+1)ⁿ, where n is a positive integer, show that the ratio of dy/dx when x =1 to d²y/dx² =1:n.

 

 

Thank you!

Answer 1

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