#### Please Solve R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 1 Sub Question 2 Maths Textbook Solution.

$The \: slope \: of \: the \: tangent \: is \frac{1}{6}$

$\text { The slope of the normal is }-6$

Hint:

$\text { The slope of tangent is the limit of } \frac{\Delta y}{\Delta x} \text { as } \Delta x \text { approaches zero. }$

$\text { The slope of normal is the normal to a curve at } P=(x, y) \text { is a line perpendicular to the tangent }$

$\text { at } P \text { and passing through } P \text { . }$

Given: $y=\sqrt{x} \text { at } x=9$

Solution:

$\text { First we have to find } \frac{d y}{d x} \text { of given function, }$

$f(x) \text { that is to find the derivative of } f(x)$

$y=\sqrt{x}$

\begin{aligned} &\therefore \sqrt[n]{x}=x^{\frac{1}{n}} \\ &y=(x)^{\frac{1}{2}} \\ &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \end{aligned}

$\text { We know that the slope of the tangent is } \frac{d y}{d x}$

\begin{aligned} &\frac{d y}{d x} \Rightarrow \frac{1}{2}(x)^{\frac{1}{2}-1} \\ &\frac{d y}{d x} \Rightarrow \frac{1}{2}(x)^{-\frac{1}{2}} \end{aligned}

Since,$x=9$

\begin{aligned} &\left(\frac{d y}{d x}\right)_{x-9} \Rightarrow \frac{1}{2}(9)^{\frac{-1}{2}} \\ &\left(\frac{d y}{d x}\right)_{x-9} \Rightarrow \frac{1}{2} \times \frac{1}{\sqrt{9}} \\ &\left(\frac{d y}{d x}\right)_{x-9} \Rightarrow \frac{1}{2} \times \frac{1}{3} \\ &\left(\frac{d y}{d x}\right)_{x-9} \Rightarrow \frac{1}{6} \end{aligned}

$\text { The slope of the tangent at } x=\text { is } \frac{1}{6}$

$\text { The slope of the normal }=\frac{-1}{\text { The slope of the tangent }}$

$\text { The slope of the normal }=\frac{-1}{\left(\frac{d y}{d x}\right)_{x-9}}$

$\text { The slope of the normal }=\frac{-1}{\frac{1}{6}}=-6$