Question #6bfa5

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1 Answer
yumean1119
Oct 23, 2017

(a) #dy/dx=-(3x^2+5y)/(5x+3y^2)#
(b) #y=-37/27x-19/9#

Explanation:

(a) Differentiate both sides of the equation.
Take notice that #(d)/(dx)y^3# = #d/(dy)y^3*((dy)/(dx))=3y^2(dy)/(dx)# (Chain Rule)
and #d/(dx)xy# = #y+x(dy/dx)# (Product Rule).

#d/(dx)(x^3+y^3+5xy)=d/dx65#
#3x^2+3y^2(dy)/(dx) + 5(y+x(dy)/dx)=0#
#(5x+3y^2)dy/dx=-3x^2-5y#

#dy/dx=-(3x^2+5y)/(5x+3y^2)#.

(b) Substitute #x=3,y=2# to #dy/dx=-(3x^2+5y)/(5x+3y^2)#.
The slope of the tangent line is #dy/dx=-(3*3^2+5*2)/(5*3+3*2^2)=-37/27#.

Therefore, the tangent line is:
#y-2=-37/27(x-3)#
#y=-37/27x-19/9#.

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