Answer
$m_{a} = \frac{1}{2}$
$m_{b}= 3$
Work Step by Step
The slope of the tangent line at a given point is equivalent to the derivative of the line evaluated at that point.
$a.) y = x^{\frac{1}{2}}$
Use the power rule to differentiate:
$y' = \frac{1}{2}x^{-\frac{1}{2}}$
Now since the point we are evaluating is at one, 1 to any exponent is itself. Leaving us with $\frac{1}{2}$ as the slope of the tangent line at $(1,1)$.
$y' = \frac{1}{2}1^{-\frac{1}{2}}$
$y' = \frac{1}{2}$
Now do the same thing with the other function.
$b.) y = x^3$
$y' = 3x^2$
$y' = 3 (1)^2$
$y' = 3$
