Do it step-by-step using implicit differentiation
y' means here dy/dx
step 1. diffrentiate both sides
1.1 d(xy +y^3)/dx = d (1)/dx
1.2 y + x d(y)/dx + 3y^2 d(y)/dx = 0 using power rule, composition rule and constant rule
1.3 ( x+ 3y^2) dy/dx = -y
1.4 Hence y' = -y / ( x+ 3y^2)
step 2.From here you can find y", the second derivative of y with respect to x.
Just differentiate both sides. I choose 1.3, but you can also use 1.4 and use Quotient Rule
2.1 (1+6yy' ) y' + (x+3y^2) y'' = -y' Using multiplication rule on the left side.
2.2 Conclusion
y " = -[ 2 y' + +6y(y'^2) ] / ( x + 3y^2) where y' = - y/ ( x+ 3y^2)
No need to replace y' by its value in terms of x and y if no-one asks for it.
I put in place a website dealing with AP-calculus, International Baccalaureate, ACT, College Physics. It will be online by the end of June 2012
By AGLAB.
Contact us @ aglab@live.com
