Given:
\(y=\sqrt{x}\) at x = 9
First, we have to find \(\frac{dy}{dx}\) of given function, f(x),i.e, to find the derivative of f(x)
⇒ \(y=\sqrt{x}\)
\(\therefore\sqrt[n]{x}=x^{\frac{1}{n}}\)
⇒ y = \((x)^{\frac{1}{2}}\)
\(\therefore \frac{dy}{dx}(x^n)\)\(=n.x^{n-1}\)
The Slope of the tangent is\(\frac{dy}{dx}\)
\(\therefore\)The Slope of the tangent at x = 9 is\(\frac{1}{6}\)
⇒ The Slope of the normal =\(\frac{-1}{\text{The Slope of the tanget}}\)
⇒ The Slope of the normal =\(\frac{-1}{(\frac{dy}{dx}x=9)}\)
⇒ The Slope of the normal =\(\frac{-1}{\frac{1}{6}}\)
⇒ The Slope of the normal = – 6
