#### Please Solve R.D. Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 1 Sub Question 3 Maths Textbook Solution.

$\text { The slope of the tangent is } 11$

$\text { The slope of the normal is } \frac{-1}{11}$

Hint:

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

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

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

Given: $y=x^{3}-x \text { at } x=2$

Solution:

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

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

$y=x^{3}-x$

$\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}$

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

\begin{aligned} &\frac{d y}{d x}=3 x^{3-1}-1 x^{1-1} \\ &\frac{d y}{d x}=3 x^{2}-1 \quad\left\{x^{0}=1\right\} \end{aligned}

Since,$x=2$

\begin{aligned} &\left(\frac{d y}{d x}\right)_{x=2} \Rightarrow 3(2)^{2}-1 \\ &\left(\frac{d y}{d x}\right)_{x-2} \Rightarrow 3(4)-1 \\ &\left(\frac{d y}{d x}\right)_{x-2} \Rightarrow 12-1 \\ &\left(\frac{d y}{d x}\right)_{x-2} \Rightarrow 11 \end{aligned}

$\text { The slope of the tangent at } x=2 \text { is } 11$

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

$\text { The slope of the normal }=\frac{-1}{\left(\frac{d y}{d x}\right)_{x-2}}$

$\text { The slope of the normal }=\frac{-1}{11}$