How do you implicitly differentiate #15=x^2+y^2#?

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Answer 1 · 2015-11-20T22:51:58

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

#d/dx(15) =d/dx(x^2+y^2)#

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

#2y(dy)/dx=-2x# ..... See foot note!

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

#color(blue)(~~~~~~~~~~~"Foot Note" ~~~~~~~~~~~~~~~)#

#color(brown)("Let "u=y^2)#

#color(brown)((du)/dy =2y)#

#color(brown)("So "(du)/dx = (du)/dy *(dy)/dx)#