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let y=4x^2. Find the change in y, Δy when x=2 and Δx=0.1 Find the differential dy when x=2 and dx=0.1
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8xdx=dy x=2 you get 16(0.1)=1.6
first one: if x = 2 then y = 16 and if x = 2.1 then y = 17.64 so \[\delta y =17.64-16=1.64\] yes?
bnut has second one
oh, I see what I was doing. Thanks guys
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