Ask a New Question
Menu

Find the equation of the tangent line to the curve y=5xcosx at the point (pi,–5pi).

The equation of this tangent line can be written in the form y=mx+b where

m=

and b=

1 answer

y=5x cos x
y'=slope m= 5cosx-5xsinx at x=pi
y'=m=-5

line equation: y= -5x+b
now to find b, you know the point(PI,-5PI) is on it, so

-5pi--5PI+b so b=0
Similar Questions
    1. answers icon 5 answers
  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 0 answers
more similar questions