Name: Constant Factor Rule in Differentiation
Uploaded: 2019-07-19
Duration: 7 min 6 s
Description: How do we find the derivative of a function having a constant multiplied with it? Is there some rule for it? Jump into this video and find out.

Question

The diagram shows part of the curve $y = 2 - \frac{18}{2 x + 3}$ , which crosses the $x -$ axis at $A$ and the $y -$ axis at $B$ .

The normal to the curve at $A$ crosses the $y -$ axis at $C$ .

Show that the equation of the line $A C$ is $9 x + 4 y = 27$ .

Accepted Answer

Given, the diagram shows part of the curve $y = 2 - \frac{18}{2 x + 3}$ , which crosses the $x -$ axis at $A$ and the $y -$ axis at $B$ .

Find the points $A$ and $B$ .

For point $C$ we have $y = 0$ .

Hence $2 - \frac{18}{2 x + 3} = 0 \Rightarrow 2 = \frac{18}{2 x + 3} \Rightarrow 2 x + 3 = 9 \Rightarrow x = 3$

For point $B$ we have $x = 0$ and $y = 2 - \frac{18}{0 + 3} \Rightarrow y = - 4$

We found coordinates $A (3, 0)$ and $B (0, - 4)$

Given, $y = 2 - \frac{18}{2 x + 3}$

Let $u = 2 x + 3$ so $y = 2 - \frac{18}{u} = 2 - 18 u^{- 1}$

$\frac{d u}{d x} = 2 (1) + 0 = 2$ and $\frac{d y}{d u} = 0 - 18 [- 1 u^{- 2}] = \frac{18}{u^{2}} = \frac{18}{{(2 x + 3)}^{2}}$

$∴ \frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$ [Use the chain rule]

$\Rightarrow \frac{d y}{d x} = \frac{18}{{(2 x + 3)}^{2}} \times 2$

$\Rightarrow \frac{d y}{d x} = \frac{36}{{(2 x + 3)}^{2}}$

When $x = 3, \frac{d y}{d x} = \frac{36}{{(6 + 3)}^{2}} = \frac{36}{81} = \frac{4}{9}$

Normal at $A :$ passes through the point $(3, 0)$ and gradient $= - \frac{9}{4}$

$y - 0 = - \frac{9}{4} (x - 3)$

$\Rightarrow 4 y = - 9 x + 27$

$\Rightarrow 9 x + 4 y = 27$

Hence, the equation of the line $A C$ is $9 x + 4 y = 27$ .

The diagram shows part of the curve $y = 2 - \frac{18}{2 x + 3}$ , which crosses the $x -$ axis at $A$ and the $y -$ axis at $B$ .

The normal to the curve at $A$ crosses the $y -$ axis at $C$ .

Show that the equation of the line $A C$ is $9 x + 4 y = 27$ .

Important Questions on Differentiation

Learn and Prepare for any exam you want !

The diagram shows part of the curve y=2-182x+3, which crosses the x-axis at A and the y-axis at B. The normal to the curve at A crosses the y-axis at C. Show that the equation of the line AC is 9x+4y=27.

Important Questions on Differentiation

Learn and Prepare for any exam you want !

The diagram shows part of the curve $y = 2 - \frac{18}{2 x + 3}$ , which crosses the $x -$ axis at $A$ and the $y -$ axis at $B$ .

The normal to the curve at $A$ crosses the $y -$ axis at $C$ .

Show that the equation of the line $A C$ is $9 x + 4 y = 27$ .