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