How do you find #(dy)/(dx)# given #x^(2/3)+y^(2/3)=pi^(2/3)#?

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Answer 1 · 2017-01-13T08:58:49

#(dy)/(dx)=(x/y)^(1/3)#

#x^(2/3)+y^(2/3)=pi^(2/3)#

differentiate wrt #x#

#d/(dx)(x^(2/3)+y^(2/3))=d/(dx)(pi^(2/3))#

the RHS #=0# since #pi# is a constant.

using the power rule, and chain rule on the #y# term

#2/3x^(-1/3)+2/3y^(-1/3)(dy)/(dx)=0#

#cancel(2/3)y^(-1/3)(dy)/(dx)=cancel(2/3)x^(-1/3)#

#(dy)/(dx)=(x^(-1/3)/(y^(-1/3)))#

#(dy)/(dx)=(y^(1/3)/(x^(1/3)))=(x/y)^(1/3)#