Question

find an equation of the tangent line to the curve at the given point. y=sqrt(x) (1,1) Ive never been good with these equations and the square root is confusing me.

Answer 1

Equation of a tangent line of a curve is:
y- y0 = f´ (x0) (x - x0 ). In this case: x0=1, y0=1
f´(x)=
`(1)/(2 √(x) )`(Derivation)
f´(x0)=
`(1)/(2)`
y - 1 = 1/2 ( x - 1 )
y - 1 = 1/2 x - 1/2
y = 1/2 x - 1/2 + 1
y =
`(1)/(2)x+ (1)/(2)`