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Find the slope of the tangent to the curve $y=\frac{1}{x}$ at the point $P\left(-2,-\frac{1}{2}\right)$

Answer

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Answer

$\boxed{-\frac{1}{4}}$ is the slope of the tangent to the curve y=$\frac{1}{x}$ at the point P$\left(-2,-\frac{1}{2}\right)$.

Steps

Step 1 ：Find the derivative of the curve y=$\frac{1}{x}$, which represents the slope of the tangent at any point on the curve.

Step 2 ：The derivative of y=$\frac{1}{x}$ is y'=$-\frac{1}{x^2}$.

Step 3 ：Substitute the x-coordinate of point P, -2, into the derivative to find the slope of the tangent at P: $m = -\frac{1}{(-2)^2} = -\frac{1}{4}$.

Step 4 ：$\boxed{-\frac{1}{4}}$ is the slope of the tangent to the curve y=$\frac{1}{x}$ at the point P$\left(-2,-\frac{1}{2}\right)$.

and ApplicationsFind the slope of the tangent to the curve $y=\frac{1}{x}$ at the point $P\left(-2,-\frac{1}{2}\right)$

Answer

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and Applications
Find the slope of the tangent to the curve $y=\frac{1}{x}$ at the point $P\left(-2,-\frac{1}{2}\right)$