How do you find #dy/dx# by implicit differentiation of #xy=4# and evaluate at point (-4,-1)?

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Answer 1 · 2016-11-17T15:09:43

#dy/dx=-1/4#

#xy=4#

Let's differentiate both sides

#xdy/dx+y=0#

#dy/dx=-y/x#

As, #y=4/x#

#dy/dx=-4/x^2#

At the point #(-4,-1)#

#dy/dx=-4/16=-1/4#

A tangent at #(-4,-1)# is

#y+1=-1/4(x+1)#

#y=-x/4-2#

graph{(y+2+x/4)(y-4/x)=0 [-17.28, 2.73, -4.94, 5.06]}