24.) Let P (a,b) be a point on the curve X^1/2 + y ^1/2 = 1. Show that the slope of the tangent at P is - b/a^1/2

24.) Let P (a,b) be a point on the curve X^1/2 + y ^1/2 = 1. Show that the slope of the tangent at P is - b/a^1/2
Use implicit differentiation:
$\sqrt{x} + \sqrt{y} = 1$

So taking the derivative:
$\frac{1}{2\sqrt{x}} + \frac{y^{\prime}}{2 \sqrt{y}} = 0$

$\frac{y^{\prime}}{2 \sqrt{y}} = -\frac{1}{2\sqrt{x}}$

$y^{\prime} = -\frac{\sqrt{y}}{\sqrt{x}}$

So at the point (x, y) = (a, b):

$y^{\prime} = -\frac{\sqrt{b}}{\sqrt{a}}$

-Dan

24) Let $P(a,b)$ be a point on the curve: . $x^{\frac{1}{2}}+ y^{\frac{1}{2}}\;=\;1$
Show that the slope of the tangent at $P$ is: . $-\left(\frac{b}{a}\right)^{\frac{1}{2}}$
Differentiate implicitly: . $\frac{1}{2}\!\cdot\!x^{-\frac{1}{2}} + \frac{1}{2}\!\cdot\!y^{-\frac{1}{2}}\left(\frac{dy}{dx}\right) \;=\;0\quad\Rightarrow\quad\frac{dy}{dx}\;=\;-\frac{y^{\frac{1}{2}}}{x^{\frac{1}{2}}} \;=\;-\left(\frac{y}{x}\right)^{\frac{1}{2}}$
At $P(a,b)\!:\;\;\frac{dy}{dx}\;=\;\left(\frac{b}{a}\r ight)^{\frac{1}{2}}$