What is the tangent of y= 1/(x^2),(1,1) ?

The tangent of y= 1/(x^2),(1,1) is y=-2x+3

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tangent of y= 1/(x^2),(1,1)

tangent of

y = \frac{1}{x ^{2}}, (1, 1)

y = - 2 x + 3

Solution steps

Find the slope of

y = \frac{1}{x ^{2}} : \frac{d y}{d x} = - \frac{2}{x ^{3}}

Compute the slope of

y = \frac{1}{x ^{2}}

(1, 1) : m = - 2

Find the line with slope m=

- 2

and passing through

(1, 1) : y = - 2 x + 3

y = - 2 x + 3

\int z^{n} ln (z) d z

\frac{\partial}{\partial x} (e^{x + y} sin (x - y))

\frac{d}{d x} (\frac{2}{3} (e^{- 2 x} + e^{x}))

\frac{x ^{4}}{2 - x ^{3}}

\frac{d}{d x} (47 x^{3} - 75 x)

What is the tangent of y= 1/(x^2),(1,1) ?
The tangent of y= 1/(x^2),(1,1) is y=-2x+3