Calculus Calculus Math Forum

 February 16th, 2016, 04:24 PM #1 Newbie   Joined: Feb 2016 From: Fort Worth Posts: 8 Thanks: 0 Please Help with d2y/dx2 What is d2y/dx2 of y^4=x^5? I have tried everything, every calculator or website gives me a different answer. Please help, thanks.
 February 16th, 2016, 04:33 PM #2 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond $\displaystyle y=x^{5/4},\quad x\ge0$ Can you continue?
 February 16th, 2016, 04:34 PM #3 Global Moderator   Joined: Dec 2006 Posts: 20,968 Thanks: 2217 Can you give at least one of your attempts, showing your working?
 February 16th, 2016, 04:59 PM #4 Newbie   Joined: Feb 2016 From: Fort Worth Posts: 8 Thanks: 0 y^4=x^5 4y^3y'=5x^4 y'=(5x^4)/(4y^3)=(5x^4*xy)/(4y^3*xy)=5y/4x * x^5/y^4 = 5y/4x y'' of 5y/4x=(4x(5y')-5y(4))/12x^2 = (4x(5-(5y/4x))-20y)/12x^2 =15y/12x^2 = 15x^3/12y^2 This is the last exercise of my webassign, I am trying to solve since 12.30pm and cant find the answer. Have tried multiple ways according to the sites i visited but all answer are incorrect. Dont really know what i am doing wrong.
 February 16th, 2016, 05:35 PM #5 Newbie   Joined: Feb 2016 From: Fort Worth Posts: 8 Thanks: 0 Got it! (5x^3)/(16y^3)
 February 16th, 2016, 05:57 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,968 Thanks: 2217 You frequently omitted required parentheses to enclose the denominator. You gave 12x^2 instead of (16x^2) in your third and subsequent lines. In the next line, the first "-" shouldn't be present. In the next line, 15 should be 5. Hence you should have obtained 5y/(16x^2), which equals (5x^3)/(16y^3) (but this last step isn't needed).

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