Evaluate dy/dt for the function at the point. x3 + y3 = 9; dx/dt = -3, x = 1, y = 2  Differentiating with respect to t on both sides;  d(x^3 + y^3)/dt = d(9)/dt dx^3/dt + d y^3/dt = 0 Applying chain rule; (dx^3/dx)*(dx/dt)  +  (dy^3/dy)*(dy/dt) =0; (3x^2)* (-3) + (3y^2)*(dy/dt) = 0 (3*1) * (-3) + (3* 4) * (dy/dt) = 0 -9 + 12 * (dy/dt) = 0 12(dy/dt) = 9 dy/dt =9/12 = 3/4 I hope it helps.... ("^' means "to the power") 

syal · Nov 26, 2009 1:58 AM

thanks a lot...didnt expect that quick but,,i really  do appreciate that.. you must be very good at maths... well man,,,i suck at maths...and i am taking this online class....i am really struggling to at least pass this course.......... well,,thanks again...hope you dont mind if have to ask another one some other time....:)

narnia · Nov 26, 2009 2:16 AM

hoina class nai jaane garya chaina ni kya ho...yeti sajilo pani na-aayeko....this is a very basic problem from differetiation... If you are skipping classes...pls dont....you are the one who pays for the college...not your boring professor..

jptsamachar · Nov 26, 2009 8:56 AM

from next time try, http://www.wolframalpha.com/

pyaradeshbasiharu · Nov 26, 2009 9:00 AM