Ask a New Question
Menu

Find the slope of the tangent line to the curve

(3x+3y−27)^(1/2)+(2xy−39)^(1/2)=8

at the point (8,4).

1 answer

√(3x+3y-27) + √(2xy-39) = 8

(3+3y')/2√(3x+3y-27) + (2y + 2xy')/2√(2xy-39) = 0

At (8,4)

(3+3y')/2√9 + (8+16y')/2√25 = 0

(1+y')/2 + (4+8y')/5 = 0

5(1+y') + 2(4+8y') = 0

5 + 5y' + 8 + 16y' = 0
21y' = -13
y' = -13/21
Similar Questions
  1. original curve: 2y^3+6(x^2)y-12x^2+6y=1dy/dx=(4x-2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the
    1. answers icon 1 answer
  2. Consider the curve defined by 2y^3+6X^2(y)- 12x^2 +6y=1 .a. Show that dy/dx= (4x-2xy)/(x^2+y^2+1) b. Write an equation of each
    1. answers icon 3 answers
  3. original curve: 2y^3+6(x^2)y-12x^2+6y=1dy/dx=(4x-2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the
    1. answers icon 0 answers
  4. original curve: 2y^3+6(x^2)y-12x^2+6y=1dy/dx=(4x-2xy)/(x^2+y^2+1) a) write an equation of each horizontal tangent line to the
    1. answers icon 3 answers
more similar questions