How do you use implicit differentiation to find #(dy)/(dx)# given #5y^2=2x^3-5y#?

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Answer 1 · 2017-08-03T12:19:43

# dy/dx=(6x^2)/(2y+1).#

#5y^2=2x^3-5y rArr 5y^2+5y=5(y^2+y)=2x^3.#

#:. d/dx{5(y^2+y)}=d/dx{2x^3}.#

#:. 5d/dx{y^2+y}=2d/dx{x^3}=2{3x^2}=6x^2...(1),# whereas,

using the Chain Rule,

#d/dx{y^2+y}=d/dx{y^2}+d/dx{y},#

#=d/dy{y^2}*d/dx{y}+dy/dx,#

#=2y*dy/dx+dy/dx.#

Utilising this in #(1),# we get,

# 5{2y*dy/dx+dy/dx}=5(2y+1)dy/dx=6x^2, # giving,

# dy/dx=(6x^2)/{5(2y+1)}#.