anonymous
  • anonymous
use the implicit differentiation to find the equation of the tangent line to the curve xy^3+xy=10at point (5,1)
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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amistre64
  • amistre64
what do you know about implicit differentation?
anonymous
  • anonymous
differentiate it and subs the point (5,1) . am i right? @amistre
amistre64
  • amistre64
yep, thats right think

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amistre64
  • amistre64
differentiate like normal, just keep your x' and y' in the problem...
anonymous
  • anonymous
y-1=-(x-5)/10
amistre64
  • amistre64
D(xy^3) + D(xy) = D(10) product rule ... product rule....constant rule
anonymous
  • anonymous
simplify and differentiate, 10y-10=-x-5 => x+10y=5now try to differentiate...will that do @amistre?
amistre64
  • amistre64
you CAN try to seperate it all, but its really just easier to work the problam as it...
amistre64
  • amistre64
(x'y + xy') + (x'y + xy') = D(10) = 0
amistre64
  • amistre64
x'(y^3) + x(3y^2)y' + x'y + xy' = 0
anonymous
  • anonymous
yeah this is fine even..whatever u find easy diddy :)
amistre64
  • amistre64
x' = dx/dx = 1 like any good fraction... now solve for y'
anonymous
  • anonymous
hey catch u later ..gotta go :) bye all
amistre64
  • amistre64
Ciao think :)
anonymous
  • anonymous
thanks @ thinker
anonymous
  • anonymous
ur welcome ..@diddy, all the best and ciao amistre...ttyl maybe after 2 days :)
amistre64
  • amistre64
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
anonymous
  • anonymous
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
amistre64
  • amistre64
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?
anonymous
  • anonymous
You didn't :)
amistre64
  • amistre64
yay!!... lately, thats an accomplishment :)
anonymous
  • anonymous
yea i think that's right, make sense now thanks :)
amistre64
  • amistre64
now plug in your x and y values to determine the slope of the tangent line at that point :)
anonymous
  • anonymous
ook got it
amistre64
  • amistre64
good job :)
anonymous
  • anonymous
YAY i got it!!! thanks u're the best :)
amistre64
  • amistre64
now, now... youre only saying that cause its true lol :)
anonymous
  • anonymous
lol
anonymous
  • anonymous
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).
anonymous
  • anonymous
never mind, got it!!!

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