^{2}+1)

^{n}, where n is a positive integer, show that the ratio of dy/dx when x =1 to d

^{2}y/dx

^{2}=1:n.

Given that y= (x^{2}+1)^{n}, where n is a positive integer, show that the ratio of dy/dx when x =1 to d^{2}y/dx^{2} =1:n.

Thank you!

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Pete R. | Experienced Math TutorExperienced Math Tutor

dy/dx = (chain rule) 2nx(x^{2}+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}]

d^{2}y/dx^{2}=(product and chain rule) 2nx(2x)(n-1)(x^{2}+1)^{n-2}^{ }+^{
}2n(x^{2}+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 = 4n^{2} - 4n + 4n = 4n^{2}

SO, when x-1, d^{2}y/dx^{2} = **4n**^{2}(2)^{n-2}

Now, the ratio of dy/dx (when x=1) to d^{2}y/dx^{2 }(when x=1) is

4n(2)^{n-2}

___________

4n^{2}(2)^{n-2}

which equals (drum roll) 1:n

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