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# Solution for Find the Slope of the Tangent and the Normal to the Following Curve at the Indicted Point Y = √ X at X = 9 ? - CBSE (Science) Class 12 - Mathematics

#### 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

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

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

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

$\text { Now },$

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

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

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Solution for question: 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. For the courses CBSE (Science), PUC Karnataka Science, CBSE (Arts), CBSE (Commerce)
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