Find the gradient of the normal to the curve y = \frac{x + 1}{x - 1} at the point where x … — JAMB UTME

Find the gradient of the normal to the curve $y = \frac{x + 1}{x - 1}$ at the point where $x = 3$ .

Answer: 2

The gradient of the normal to the curve at

x = 3

2

Differentiating

y = \frac{x + 1}{x - 1}

via the quotient rule gives

\frac{dy}{dx} = \frac{-2}{(x - 1)^2}

. At

x = 3

, the tangent gradient is

m_t = \frac{-2}{4} = -\frac{1}{2}

. Since the normal is perpendicular to the tangent, its gradient is

m_n = -\frac{1}{m_t} = 2

Differentiate the rational function with respect to

x

Applying the quotient rule

\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v u' - u v'}{v^2}

where

u = x + 1

and

v = x - 1

yields

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

The first derivative determines the slope function of the tangent line to the curve.

Substitute

x = 3

into the derivative to find the tangent slope

m_t

m_t = \frac{-2}{(3 - 1)^2} = \frac{-2}{4} = -\frac{1}{2}

Evaluating the derivative at the given

x

-coordinate provides the exact gradient of the tangent at that point.

Calculate the slope of the normal line

m_n

m_n = -\frac{1}{m_t} = -\frac{1}{-1/2} = 2

The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal of the tangent slope.

The slope of the normal line to a curve at a given point is the negative reciprocal of the slope of the tangent line at that point (

m_n = -\frac{1}{m_t}

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