User:Byron Vickers/latex

The equation to the tangent of $$ f(x) = {\sqrt{x}} $$ at the point (4,2) is:

$$ y-y_1 = M^T (x-x_1)$$

$$ M^T=f'(x)={\frac{1}{2}}x^{{\frac{1}{2}}-1}={\frac{1}{2{\sqrt{x}}}} $$