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anonymous
 5 years ago
use the implicit differentiation to find the equation of the tangent line to the curve xy^3+xy=10at point (5,1)
anonymous
 5 years ago
use the implicit differentiation to find the equation of the tangent line to the curve xy^3+xy=10at point (5,1)

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0what do you know about implicit differentation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0differentiate it and subs the point (5,1) . am i right? @amistre

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yep, thats right think

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0differentiate like normal, just keep your x' and y' in the problem...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0D(xy^3) + D(xy) = D(10) product rule ... product rule....constant rule

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0simplify and differentiate, 10y10=x5 => x+10y=5now try to differentiate...will that do @amistre?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0you CAN try to seperate it all, but its really just easier to work the problam as it...

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0(x'y + xy') + (x'y + xy') = D(10) = 0

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x'(y^3) + x(3y^2)y' + x'y + xy' = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah this is fine even..whatever u find easy diddy :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0x' = dx/dx = 1 like any good fraction... now solve for y'

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey catch u later ..gotta go :) bye all

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ur welcome ..@diddy, all the best and ciao amistre...ttyl maybe after 2 days :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0Its 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oki 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

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0Let 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?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0yay!!... lately, thats an accomplishment :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yea i think that's right, make sense now thanks :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0now plug in your x and y values to determine the slope of the tangent line at that point :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0YAY i got it!!! thanks u're the best :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0now, now... youre only saying that cause its true lol :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i'm having some trouble with this one... Find the slope of the tangent line to the curve 1x^22xy3y^3=185 at the point (1,4).

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0never mind, got it!!!
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