How do you implicitly differentiate #x^4+2y^2=8 #?

1 Answer
Alan P.
Nov 20, 2015

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

Explanation:

#(dx^4)/(dx) = 4x^3#

#(d2y^2)/(dx) = (d2y^2)/(dy)*(dy)/(dx) = 4y(dy)/(dx)#

#(d (x^4+2y^2))/(dx) = (dx^4)/(dx)+(d2y^2)/(dx) = 4x^3+4y(dy)/(dx)#

#(d8)/(dx) = 0#

Since #x^4+2y^2=8#
#color(white)("XXX")(d(x^4+2y^2))/(dx) = (d8)/(dx)#

#color(white)("XXX")4x^3+4y(dy)/(dx)=0#

#color(white)("XXX")4y(dy)/(dx) = -4x^3#

#color(white)("XXX")(dy)/(dx) = (-4x^3)/(4y)=-x^3/y#

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