Question

How do you find an equation of the tangent line to the curve at the given point `y = 2xcosx` and `(pi, -2pi)`?

Answer 1

`2x+y=0.`

Explanation:

`y=2xcosx rArr dy/dx=2{x(cosx)'+(cosx)(x)'}` [Product Rule for Diff.]
`:. dy/dx=2(-xsinx+cosx)`

`:.[dy/dx]_(x=pi,y=-2pi)=2(-pisinpi+cospi)=-2.`

This means that the slope m of tangent (tgt.) line to the given curve is `m=-2.`

Tgt. line passes thro. the pt. `(pi,-2pi).`

Therefore, its eqn. is `y-(-2pi)=(-2)(x-pi),`

`i.e., y+2pi+2x-2pi=0,` or, `2x+y=0.`