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what do you know about implicit differentation?
differentiate it and subs the point (5,1) . am i right? @amistre
yep, thats right think
differentiate like normal, just keep your x' and y' in the problem...
D(xy^3) + D(xy) = D(10) product rule ... product rule....constant rule
simplify and differentiate, 10y-10=-x-5 => x+10y=5now try to differentiate...will that do @amistre?
you CAN try to seperate it all, but its really just easier to work the problam as it...
(x'y + xy') + (x'y + xy') = D(10) = 0
x'(y^3) + x(3y^2)y' + x'y + xy' = 0
yeah this is fine even..whatever u find easy diddy :)
x' = dx/dx = 1 like any good fraction... now solve for y'
hey catch u later ..gotta go :) bye all
Ciao think :)
thanks @ thinker
ur welcome ..@diddy, all the best and ciao amistre...ttyl maybe after 2 days :)
Its important to realize that there is nothing that special when doing implicits.. for example: D(x^4) = x' 4x^3 D(3y^2) = y' 6y D(3xy) = 3xy' + x'3y D(y) = y' D(x) = x' its all normal rules
oki so i used this form amistre x'(y^3) + x(3y^2)y' + x'y + xy' = 0 and i came up with 2y^3/3xy^2+1, but i'm not sure that's right
Let me double check the work... x'(y^3) + x(3y^2)y' + x'y + xy' = 0 y^3 + y'(3xy^2) + y + y'x = 0 y'(3xy^2) + y'x = -y^3 - y y'(3xy^2 + x) = -y^3 -y -y^3 -y y' = ----------- 3xy^2 + x did I make any mistakes in this?
You didn't :)
yay!!... lately, thats an accomplishment :)
yea i think that's right, make sense now thanks :)
now plug in your x and y values to determine the slope of the tangent line at that point :)
ook got it
good job :)
YAY i got it!!! thanks u're the best :)
now, now... youre only saying that cause its true lol :)
i'm having some trouble with this one... Find the slope of the tangent line to the curve -1x^2-2xy-3y^3=-185 at the point (-1,4).
never mind, got it!!!