#### Question

Find the slope of the tangent and the normal to the following curve at the indicted point \[y = \sqrt{x} \text { at }x = 9\] ?

#### Solution

\[ y = \sqrt{x} = x^\frac{1}{2} \]

\[ \Rightarrow \frac{dy}{dx} = \frac{1}{2} x^\frac{- 1}{2} = \frac{1}{2\sqrt{x}}\]

\[\text { When }x=9, y = \sqrt{x} = \sqrt{9} = 3\]

\[\text { Now }, \]

\[\text { Slope of the tangent }= \left( \frac{dy}{dx} \right)_\left( 9, 3 \right) =\frac{1}{2\sqrt{9}}=\frac{1}{6}\]

\[\text { Slope of the normal }=\frac{- 1}{\left( \frac{dy}{dx} \right)_\left( 9, 3 \right)}=\frac{- 1}{\left( \frac{1}{6} \right)}=-6\]

Is there an error in this question or solution?

Solution Find the Slope of the Tangent and the Normal to the Following Curve at the Indicted Point Y = √ X at X = 9 ? Concept: Tangents and Normals.