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)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "\n" ); document.write( "— \n" ); document.write( "For the first part we can use u-substitution: \n" ); document.write( "Let \n" ); document.write( " \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \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 \n" ); document.write( "— \n" ); document.write( " \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 \n" ); document.write( "\n" ); document.write( "—\r \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "Which can be re-written after much algebra: \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "and that factors (thanks to an online factoring tool) to: \n" ); document.write( " |