Question

Calculas: Any and all help is appreciated!!!!!?
Consider the curve given by x2 + sin(xy) + 3y2 = C, where C is a constant. The point (1, 1) lies on this curve. Use the tangent line approximation to approximate the y-coordinate when x = 1.01.

0.996
1
1.004
Cannot be determined
1.388

Answer 1

calculas? i've never heard of it

Answer 2

i joke.

Answer 3

can u help?????

Answer 4

the tangent line is the derivative
find the first derivative of x2 + sin(xy) + 3y2 = C

Answer 5

1.004?????

Answer 6

hold on

Answer 7

find the first derivative and then solve for y when x = 1.01, that's what they're asking right?...

Answer 8

d/dx(x^2+sin(xy) + 3y2=((2)(x)^(2-1))+sin'(xy)+(2)(3)y^(2-1))+????? is that right?

Answer 9

@Isaiah.Feynman feel free to butt in , i suck at calc >.>

Answer 10

y cos(xy)+2x

Answer 11

2x + cos (xy) +6yy' = 0
i think you're supposed to use the product rule for sin (xy) tho...

Answer 12

OK

Answer 13

So...

Answer 14

\[\huge \frac{ d }{ dx } \sin(xy)\]
..

Answer 15

(2x)+(cos(xy))+(6y)

Answer 16

2(1.01)+cos(1.01)+6y=0

Answer 17

2(1.01)+cos(1.01)y+6y=0

Answer 18

2(1.01)+0.531861y+6y=0

Answer 19

6.531861y=2.02

Answer 20

y=(6.531861/2.02)

Answer 21

y=3.2335945544554455445544554455445544554455445544554455

Answer 22

idk