Find the slope of the tangent to the curve y= | vedclass

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Question

(A) The given curve is $y = \frac{x-1}{x-2}$.To find the slope of the tangent,we calculate the derivative $\frac{dy}{dx}$ using the quotient rule:$\fr

Vedclass · Accepted Answer

$\frac{-1}{64}$

Vedclass · Answer

(A) The given curve is $y = \frac{x-1}{x-2}$.To find the slope of the tangent,we calculate the derivative $\frac{dy}{dx}$ using the quotient rule:$\frac{dy}{dx} = \frac{(x-2)\frac{d}{dx}(x-1) - (x-1)\frac{d}{dx}(x-2)}{(x-2)^2}$$\frac{dy}{dx} = \frac{(x-2)(1) - (x-1)(1)}{(x-2)^2}$$\frac{dy}{dx} = \frac{x-2-x+1}{(x-2)^2} = \frac{-1}{(x-2)^2}$Now,evaluate the slope at $x=10$:$\left. \frac{dy}{dx} \right|_{x=10} = \frac{-1}{(10-2)^2} = \frac{-1}{8^2} = \frac{-1}{64}$.Thus,the slope of the tangent at $x=10$ is $\frac{-1}{64}$.

Find the slope of the tangent to the curve $y=\frac{x-1}{x-2}, x \neq 2$ at $x=10$.

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