document.write( "Question 1113571: CALCULUS. Find (d^2 y)/(dx^2) if y=(x^2 +1)^5. Please help. Thankyou in advance! \n" ); document.write( "
Algebra.Com's Answer #728637 by math_helper(2461)\"\" \"About 
You can put this solution on YOUR website!
\"+y+=+%28x%5E2%2B1%29%5E5+\"\r
\n" ); document.write( "\n" ); document.write( "—
\n" ); document.write( "For the first part we can use u-substitution:
\n" ); document.write( "Let \"+u+=+%28x%5E2%2B1%29+\"
\n" ); document.write( " \"+du+=+2x+dx+\"\r
\n" ); document.write( "\n" ); document.write( " \"+y+=+u%5E5+\"
\n" ); document.write( " \"++dy+=+5%28u%5E4%29+du++=+5%28x%5E2%2B1%29%5E4%282x+dx%29+=+10x%28%28x%5E2%2B1%29%5E4%29+dx\" —> \"+dy%2Fdx+=+10x%28x%5E2%2B1%29%5E4+\"
\n" ); document.write( "—\r
\n" ); document.write( "\n" ); document.write( "Aside:
\n" ); document.write( "Notice that the chain rule (which applies to composite functions) could have been used:
\n" ); document.write( "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.
\n" ); document.write( "—
\n" ); document.write( "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.
\n" ); document.write( "
\n" ); document.write( "Now onward to get \"d%5E2y%2Fdx%5E2+\"
\n" ); document.write( "—
\n" ); document.write( "\"+dy%2Fdx+=+10x%28x%5E2%2B1%29%5E4+\" is a product of two functions 10x and (x^2+1)^4
\n" ); document.write( "—
\n" ); document.write( "The product of functions lends itself to using the product rule:
\n" ); document.write( "The product rule is (f*g)' = fg' + f'g (where f=f(x) and f' = df/dx, g=g(x), g'=dg/dx)
\n" ); document.write( "\"The first times the derivative of the 2nd, plus the 2nd times the derivative of the first.\"
\n" ); document.write( "—\r
\n" ); document.write( "\n" ); document.write( "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%28green%29+\" below.\r
\n" ); document.write( "\n" ); document.write( "—\r
\n" ); document.write( "\n" ); document.write( "
\n" ); document.write( "\"+d%5E2y%2Fdx%5E2+=+%2880x%5E2%29%28x%5E2%2B1%29%5E3+%2B+10%28x%5E2%2B1%29%5E4+\"
\n" ); document.write( "
\n" ); document.write( "Which can be re-written after much algebra:
\n" ); document.write( "\"+d%5E2y%2Fdx%5E2+=+highlight%2890x%5E8+%2B+280x%5E6+%2B+300x%5E4+%2B+120x%5E2+%2B+10%29+\"\r
\n" ); document.write( "\n" ); document.write( "and that factors (thanks to an online factoring tool) to:
\n" ); document.write( "\"+d%5E2y%2Fdx%5E2+=+highlight%2810%28x%5E2%2B1%29%5E3%289x%5E2%2B1%29%29+\"
\n" ); document.write( "
\n" );