Determine equation of the tangent to the curve

Question

y= xsinx at the point x=pi/2?

Answer 1

y = x sinx

 

First find the y-coordinate x =π/2 then

 

y = x sinx

 

y =π/2 *  sinπ/2

 

y = π/2 * 1 = π/2

 

y = π/2

point = ( x , y ) = (π/2,π/2)

 y = x sinx

Derivative of both sides.

y '  = d /dx (x sin x )

Product rule : d/dx (uv) = (u'v + v'u)

y '  = d /dx (x) sin x + xd/dx(sin x)

y ' =  sin x + x cosx

x = π/2  so the slope of the tangent line.

y ' =  sinπ/2 + π/2cosπ/2

y ' = 1 + π/2(0) = 1

y ' = 1

m = 1

point slope formula : y - y₁ = m ( x - x ₁)

Substitute the values m = 1 and (x ₁ , y₁ ) = (π/2,π/2) in point slope formula.

y - π/2 = 1( x - π/2)

y - π/2 =  x - π/2

Add π/2 to both sides

y - π/2 + π/2 =x - π/2 + π/2

y = x

The normal line is the perpendicular line. (Opposite Reciprical Slope)

y - π/2 = - 1( x - π/2)

y - π/2 = -  x + π/2

y  = -  x + π/2+π/2

y = - x + π

y = - x + π

Tangent line : y = - x + π.