## Calculus 10th Edition

$m_{a}=-\frac{1}{2}$ $m_{b}= -1$
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: $y' = -\frac{1}{2}x^{-\frac{3}{2}}$ 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 $(1,1)$. $y' = -\frac{1}{2}1^{-\frac{3}{2}}$ $y' = -\frac{1}{2}$ Now do the same thing with the other function. $b.) y = x^{-1}$ $y' = -1x^{-2}$ $y' = -1 (1)^{-2}$ $y' = -1$