Find the equation of the tangent line to the curve y=2sinx at the point ((pi/6), 1). The equation of the tangent line can be written in y=mx+b form, where m=sqrt3.

What is b?

1 answer

Axel
Similar Questions
  1. Find an equation of the tangent line to the given curve at the given pointy=(1/cosx)-2cosx at((pi/3),1) y'=0+2sinx slope = √3
    1. answers icon 1 answer
  2. 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
  3. 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
  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