Name: Derivative of Functions in Parametric Form
Uploaded: 2019-07-19
Duration: 8 min 37 s
Description: Learn everything you'll need to know about derivatives of functions in parametric forms with this quick and easy lesson that'll cover the concept comprehensi...

Question

The diagram shows the curve $y = \sqrt{(\frac{1 - x}{1 + x})} .$ Find $\frac{d y}{d x}$ .

The gradient of the normal to the curve has its maximum value at the point $P$ shown in the diagram. Find, by differentiation, the $x$ -coordinate of $P .$

Accepted Answer

Given equation of the curve is $y = \sqrt{(\frac{1 - x}{1 + x})} . . . . . (i)$

It is given that the gradient of the normal to the curve has its maximum value at the point $P$ . We need to find the $x$ -coordinate of point $P$ .

We need to find: $\frac{d y}{d x}$ .

Let $u = \frac{1 - x}{1 + x} . . . . (i i)$

$\Rightarrow y = \sqrt{u} . . . . (i i i)$ [Using $(i)$ ]

Now, $\frac{d u}{d x} = \frac{(1 + x) \frac{d}{d x} (1 - x) - (1 - x) \frac{d}{d x} (1 + x)}{{(1 + x)}^{2}}$ [using $(i i)$ ]

$= \frac{(1 + x) \times (- 1) - (1 - x) \times (1)}{{(1 + x)}^{2}}$

$= \frac{- 1 - x - 1 + x}{{(1 + x)}^{2}}$

$= \frac{- 2}{{(x + 1)}^{2}} . . . . (i v)$

Now, $\frac{d y}{d u} = \frac{1}{2 \sqrt{u}} . . . . (v)$ [From $(i i i)$ , and applying the differentiation formula, $\frac{d}{d x} (x^{n}) = n x^{n - 1}$ ]

We know that

$\frac{d y}{d x} = \frac{d y}{d u} \times \frac{d u}{d x}$

$= \frac{1}{2 \sqrt{u}} \times \frac{- 2}{{(x + 1)}^{2}}$ [From $(i v)$ , and $(v)$ ]

$= - \frac{1}{\sqrt{u} {(x + 1)}^{2}}$

$= \frac{- 1}{(\sqrt{\frac{1 - x}{1 + x}}) \times {(x + 1)}^{2}}$

$= \frac{- 1}{{(x + 1)}^{\frac{3}{2}} \sqrt{1 - x}}$

Hence, $\frac{d y}{d x} = \frac{- 1}{{(x + 1)}^{\frac{3}{2}} \sqrt{1 - x}} . . . . . (v i)$ .

Now, the gradient of the normal to the given curve is given by $- \frac{d x}{d y}$ .

Thus, using $(v i)$ , we have

$- \frac{d x}{d y} = {(x + 1)}^{\frac{3}{2}} \sqrt{1 - x}$

Now, for the maximum point, we need to differentiate, $(- \frac{d x}{d y})$ .

Thus, $- \frac{d^{2} x}{d y^{2}} = \frac{d}{d x} [{(x + 1)}^{\frac{3}{2}} \sqrt{1 - x}]$

$= {(x + 1)}^{\frac{3}{2}} \times \frac{d}{d x} (\sqrt{1 - x}) + \sqrt{1 - x} \frac{d}{d x} [{(x + 1)}^{\frac{3}{2}}]$ $[∵ \frac{d}{d x} (u v) = u \frac{d v}{d x} + v \frac{d u}{d x}]$

$= {(x + 1)}^{\frac{3}{2}} [\frac{1}{2 \sqrt{1 - x}} \frac{d}{d x} (1 - x)] + (\sqrt{1 - x}) \times \frac{3}{2} {(x + 1)}^{\frac{3}{2} - 1}$ [Applying the chain rule]

$= {(x + 1)}^{\frac{3}{2}} [\frac{1}{2 \sqrt{1 - x}} \times (- 1)] + (\sqrt{1 - x}) \times \frac{3}{2} \sqrt{(x + 1)}$ $[∵ \frac{d}{d x} (x^{n}) = n x^{n - 1}, n \in R]$

$= {(x + 1)}^{\frac{3}{2}} [\frac{- 1}{2 \sqrt{1 - x}}] + \frac{3}{2} (\sqrt{1 - x}) \sqrt{(x + 1)}$

$= \frac{- {(x + 1)}^{\frac{3}{2}} + 3 (1 - x) \sqrt{x + 1}}{2 \sqrt{1 - x}}$

$= \frac{\sqrt{x + 1} [- (x + 1) + 3 (1 - x)]}{2 \sqrt{1 - x}}$

$= \frac{\sqrt{x + 1} [- x - 1 + 3 - 3 x]}{2 \sqrt{1 - x}}$

$= \frac{\sqrt{x + 1} [- 4 x + 2]}{2 \sqrt{1 - x}}$

Now, for the maximum point we should have $- \frac{d^{2} x}{d y^{2}} = 0$

$\Rightarrow \frac{\sqrt{x + 1} [- 4 x + 2]}{2 \sqrt{1 - x}} = 0$

$\Rightarrow x = - 1, \frac{1}{2}$

Since the given equation of the curve is $y = \sqrt{\frac{1 - x}{1 + x}}$ which is not defined for $x = - 1$ .

Hence, the gradient of the normal to the curve has its maximum value at the point $x = \frac{1}{2}$ .

Hence, the $x$ -coordinate of point $P$ is $0.5$ .

The diagram shows the curve $y = \sqrt{(\frac{1 - x}{1 + x})} .$ Find $\frac{d y}{d x}$ .

The gradient of the normal to the curve has its maximum value at the point $P$ shown in the diagram. Find, by differentiation, the $x$ -coordinate of $P .$

Important Questions on Differentiation

Learn and Prepare for any exam you want !

The diagram shows the curve y=1-x1+x. Find dydx. The gradient of the normal to the curve has its maximum value at the point P shown in the diagram. Find, by differentiation, the x-coordinate of P.

Important Questions on Differentiation

Learn and Prepare for any exam you want !

The diagram shows the curve $y = \sqrt{(\frac{1 - x}{1 + x})} .$ Find $\frac{d y}{d x}$ .

The gradient of the normal to the curve has its maximum value at the point $P$ shown in the diagram. Find, by differentiation, the $x$ -coordinate of $P .$