How do you use implicit differentiation to find the slope of the curve given #xy=-8# at (4,-2)?

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Answer 1 · 2016-08-01T16:19:16

The slope of tgt. at pt.#(4,-2)=1/2#.

Let us recall that #dy/dx# gives the slope of tgt. at the pt. #(x,y)#.

Now, #xy=-8#

#rArr d/dx(xy)=d/dx(-8)=0#.

#rArr xdy/dx+yd/dx(x)=0#.

#rArr xdy/dx+y=0#.

#rArr dy/dx=-y/x#

#:.# the slope of tgt. at pt. #(x,y)=-y/x#.

#:.# the slope of tgt. at pt.#(4,-2)=-(-2)/4=1/2#.