Question

Find the equation of the tangent line to the curve y2+4xy+x2 = 13
at the point (2; 1).

Answer 1

find the first derivative:
2y*y' + 4(x*y' + y) + 2x = 0
2y*y' + 4x*y' + 4y + 2x = 0

Answer 2

all you have to do now is find y' as it is also the slope of the tangent line. Plug in 2 for x and 1 for y:
2*y' + 4*2*y'+4*1+2*2 = 0
2y'+8y' = -4-4
10y' = -8
y' = M = -8/10

Answer 3

Now use the formula:
y - y1 = M(x-x1)
y-1 = (-4/5)(x-2)

Answer 4

M = -8/10 = -4/5 thats how i got -4/5 in the previous part..

Answer 5

Simplify:
y-1 = (-4/5)(x-2)
y = -4x/5 + 8/5 +1
y = -4x/5 + 13/5

Answer 6

Please click on become a fan if I helped, I really want to get to the next level!! Thanks =)

Answer 7

yeah that helped a lot! it's been a few years since i did this stuff so i totally forgot about using y' as an intermediate variable, makes sense now tho, thanks and I will fan you :)