The slope of the tangent

Find the slope of the tangent to the curve 6x^3+7x^2y+6xy^2+10y^3=29 at (1,1)?

Help me!!

6x^3 + 7x^2 * y + 6xy^2 + 10y^3 = 29
18x^2 * dx + 7 * (x^2 * dy + 2xy * dx) + 6 * (x * 2y * dy + y^2 * dx) + 30y^2 * dy = 0

x = 1
y = 1

18 * dx + 7 * (dy + 2dx) + 6 * (2dy + dx) + 30 * dy = 0
18 * dx + 7 * dy + 14 * dx + 12 * dy + 6 * dx + 30 * dy = 0
18 * dx + 14 * dx + 6 * dx + 7 * dy + 12 * dy + 30 * dy = 0
38 * dx + 49 * dy = 0
49 * dy = -38 * dx
dy/dx = -38 / 49

y - 1 = (-38/49) * (x - 1)
y = (-38/49) * x + 38/49 + 49/49
y = (-38/49) * x + 87/49
y = (1/49) * (87 - 38x)

do your own homework
jk.

first you need to find the derivative
remember to use product rule when needed
18x^2+14xy+7x^2(yprime)+6y^2+2y(yprime…

then you put all the terms to one side and leave the terms with yprime on the other. so...
7x^2(yprime)+2y(yprime)6x+30y^2(yprime…

factor out the yprimes
yprime(7x^2+2y6x+30y^2)=-18x^2-14xy-6y…

divide to get yprime alone
yprime=(-18x^2-14xy-6y^2)/(7x^2+2y6x+3…

as you probably learned, yprime gives you the slope of the equation. now you just plug in the (1,1) to find the slope at that point
yprime= (-18-14-6)/(7+12+30)

answer: y prime=slope=-38/49

this is kind of an odd slope so you might wanna check to see if the original equation is right

You would take the derivative to find y' and then plug 1 into x and 1 into y of that equation. Since this has x and y mixed, you will need to use implicit differentiation and then solve for dx/dy.

d/dx both sides:

d/dx(6x^3+7x^2y+6xy^2+10y^3) = dx/dy(29)

18x^2 + (14xy + 7x^2y') + (6y^2 + 12xy*y') + (30y^2y')= 0

y'(7x^2 + 12xy + 30y^2) = -18x^2 -14xy -6y^2

y' = (-18x^2 - 14xy - 6y^2)/(7x^2 + 12xy + 30y^2)

Then plug in 1 for x and 1 for y: