Question 1113571
{{{ y = (x^2+1)^5 }}}

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For the first part we can use u-substitution:
Let {{{ u = (x^2+1) }}}
     {{{ du = 2x dx }}}

   {{{ y = u^5 }}}
   {{{  dy = 5(u^4) du  = 5(x^2+1)^4(2x dx) = 10x((x^2+1)^4) dx}}} —> {{{ dy/dx = 10x(x^2+1)^4 }}}
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Aside:
Notice that the chain rule (which applies to composite functions) could have been used:
The chain rule says (f(g(x)))' = f'(g)*g')  "The derivative of f of g(x) is the derivative of f(g(x)) times the derivative of g(x)"   Often used for expressions like (mx+k)^n where, m and k are numbers, g(x)=mx+k and f(x)=x^n,  one takes  n*(mx+k)^(n-1) then multiplies by dg/dx = d(mx+k)/dx = m to arrive at  m(mx+k)^(n-1).   Think of it as an implicit use of u-substitution.  
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I mention the chain rule because I will use it for finding the 2nd derivative.  However, that step happens to be identical in procedure to how we found the first derivative.
 
Now onward to get {{{d^2y/dx^2 }}} …
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{{{ dy/dx = 10x(x^2+1)^4 }}}  is a product of two functions 10x and (x^2+1)^4
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The product of functions lends itself to using the product rule:
The product rule is  (f*g)' = fg' + f'g   (where f=f(x) and f' = df/dx, g=g(x), g'=dg/dx) 
"The first times the derivative of the 2nd, plus the 2nd times the derivative of the first."
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The first is "10x" , the 2nd is "(x^2+1)^4"  so we will do  10x * derivative of(x^2+1)^4) + (x^2+1)^4 * derivative of(10x),   noting that finding the derivative of (x^2+1)^4 wrt x is very similar to how we found the first derivative (it is the same process just different numbers).  The part I'm referring to is highlighted in {{{ green(green) }}} below.

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{{{ d^2y/dx^2 = 10x (green(4*(x^2+1)^3*(2x))) + (x^2+1)^4 * 10 }}}
{{{ d^2y/dx^2 = (80x^2)(x^2+1)^3 + 10(x^2+1)^4 }}}
   
Which can be re-written after much algebra:
{{{ d^2y/dx^2 = highlight(90x^8 + 280x^6 + 300x^4 + 120x^2 + 10) }}}

and that factors (thanks to an online factoring tool) to:
{{{ d^2y/dx^2 = highlight(10(x^2+1)^3(9x^2+1)) }}}