How do you implicitly differentiate #x^2 + y^2 = 1/2 #?

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Answer 1 · 2018-06-22T12:55:58

# dy/dx = - x/y#

# x^2+y^2 =1/2#

#d/dx( x^2+y^2) = d/dx (1/2)#

# 2 x+ 2y dy/dx = 0#

# :. 2y dy/dx = -2 x#

# :. dy/dx = -(2 x)/(2y)#

# :. dy/dx = - x/y# [Ans]

Answer 2 · 2018-06-22T13:00:05

#dy/dx=-x/y#

#color(blue)"differentiate implicitly with respect to x"#

#"noting that "d/dx(y^2)=2ydy/dx#

#2x+2ydy/dx=0#

#2ydy/dx=-2x#

#dy/dx=-(2x)/(2y)=-x/y#