Find an equation of the line tangent to the graph of y= x^2 + sin(pi/2)x at x=-1

To find the tangent to the curve y = x^2+sin(pi/2)x at x =
1.

At x= 1, y = = 1^2 +sin (pi/2)(1) = 1 +1 =
2. 

So (x,y ) = (1,2) is a point on the curve where we like
to determine the equation of the tangent to the curve.

We
know the tangent at (x1 y1) is given by:

y-y1 ={ (dy/dx) at
(x1 , y1) }(x-x1))....(1)

We differentiate y = x^2+sin
(pi/2)x.

dy/dx = 2x+
(pi/2)cos(pi/2)x 

{dy/dx at x = 1} = 2+(pi/2) cos(pi/2) =
2+(pi/2)*0 = 2.

Therefore substituting  (x1 ,y1) = (1,2)
and {dy/dx ) at x= 1) = 2 in  (1), we get the equation of the
tangent:

y - 2 = 2(x-1).

 y-2
= 2x-2 .

 y = 2x is the equation of the tangent to  y=
x^2+sin (pi/2)x at x= 1.