Question

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

Answer 1

what do you know about implicit differentation?

Answer 2

differentiate it and subs the point (5,1) . am i right? @amistre

Answer 3

yep, thats right think

Answer 4

differentiate like normal, just keep your x' and y' in the problem...

Answer 5

y-1=-(x-5)/10

Answer 6

D(xy^3) + D(xy) = D(10)

product rule ... product rule....constant rule

Answer 7

simplify and differentiate, 10y-10=-x-5 => x+10y=5now try to differentiate...will that do @amistre?

Answer 8

you CAN try to seperate it all, but its really just easier to work the problam as it...

Answer 9

(x'y + xy') + (x'y + xy') = D(10) = 0

Answer 10

x'(y^3) + x(3y^2)y' + x'y + xy' = 0

Answer 11

yeah this is fine even..whatever u find easy diddy :)

Answer 12

x' = dx/dx = 1 like any good fraction... now solve for y'

Answer 13

hey catch u later ..gotta go :) bye all

Answer 14

Ciao think :)

Answer 15

thanks @ thinker

Answer 16

ur welcome ..@diddy, all the best
and ciao amistre...ttyl maybe after 2 days :)

Answer 17

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

Answer 18

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

Answer 19

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?

Answer 20

You didn't :)

Answer 21

yay!!... lately, thats an accomplishment :)

Answer 22

yea i think that's right, make sense now
thanks :)

Answer 23

now plug in your x and y values to determine the slope of the tangent line at that point :)

Answer 24

ook got it

Answer 25

good job :)

Answer 26

YAY i got it!!! thanks u're the best :)

Answer 27

now, now... youre only saying that cause its true lol :)

Answer 28

lol

Answer 29

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).

Answer 30

never mind, got it!!!