Question

Use implicit differentiation to find the equation of the tangent line to the curve xy^3+xy=2 at the points (1,1)

Answer 1

y³ + 3xy²y' + y + xy' = 0
y' = (-y³ -y)/(3xy² + x)
Equation of tangent line:
y - 1 = y'(1,1) * (x-1)
y-1 = -1/2 * (x-1)

Answer 2

thank you
can you help me with another one

Answer 3

find the slope of the tangent line to the curve
3x^2-3xy-4y^3=-114 at points (2,3)

Answer 4

sure

Answer 5

wait actually i just got that one

Answer 6

i need help with this one
i got an answer but its telling me its wrong...

Answer 7

6x -3y - 3xy' -12y²y' = 0
y' = (-6x + 3y)/(-3x -12y²)
The value of the slope at (2,3)
y'(2,3) = 1/38

Answer 8

find the slope of the tangent line to the curve
1x^2+4xy+2y^3=-92 at points (-2,-4)

Answer 9

2x +4y + 4xy' + 6y²y' = 0
y' = (-2x -4y)/(4x + 6y²)
Value of the slope ate (-2,-4)
y'(-2,-4) = 5/22

Answer 10

thanks

Answer 11

welcome xD

Answer 12

i have one more if you dont mind helping:)

Answer 13

last one lol

Answer 14

sure

Answer 15

if 4x^2+3x+xy=1 and y(1)=-6, find y'(1) by implicit dif.

Answer 16

8x + 3 +y +xy' = 0
y' = (-8x -3 -y)/x
at point (1,-6)
y'(1) = -5

Answer 17

thanks for the help:)

Answer 18

any time xD