Asked by LEON
Use implicit differentiation to find the equation of the tangent line to the curve xy^3+2xy=9 at the point (31). The equation of this tangent line can be written in the form y=mx+b where m is
Answers
Answered by
Damon
x(3y^2dy) + y^3(dx) +2xdy + 2ydx = 0
dy (3xy^2+2x) = -dx (y^3+2y)
dy/dx = -(y^3+2y)/(3xy^2+2x)
I assume you mean the point (3,1)
x = 3
y = 1
so
m = dy/dx = -(1+2)/(9+6) = - 3/15 = -1/5
put in point
1 = m * 3 + b
1 = -3/5 + b
b = 8/5
y = -x/5 + 8/5
5y = 8-x
dy (3xy^2+2x) = -dx (y^3+2y)
dy/dx = -(y^3+2y)/(3xy^2+2x)
I assume you mean the point (3,1)
x = 3
y = 1
so
m = dy/dx = -(1+2)/(9+6) = - 3/15 = -1/5
put in point
1 = m * 3 + b
1 = -3/5 + b
b = 8/5
y = -x/5 + 8/5
5y = 8-x
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.