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Question
Find the slope of the tangent and the normal to the following curve at the indicted point y = x3 − x at x = 2 ?
Solution
\[y = x^3 - x\]
\[ \Rightarrow \frac{dy}{dx} = 3 x^2 - 1\]
When `x=2,`
`y=x^3-x`
`=2^3-2`
`=8-2`
`=6`
\[\text { Now }, \]
\[\text { Slope of the tangent }= \left( \frac{dy}{dx} \right)_\left( 2, 6 \right) =3 \left( 2 \right)^2 -1=11\]
\[\text { Slope of the normal }=\frac{- 1}{\left( \frac{dy}{dx} \right)_\left( 2, 6 \right)}=\frac{- 1}{11}\]
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