What is the tangent of y= 1/(x^2),(3, 1/9) ?

The tangent of y= 1/(x^2),(3, 1/9) is y=-2/27 x+1/3

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tangent of y= 1/(x^2),(3, 1/9)

tangent of

y = \frac{1}{x ^{2}}, (3, \frac{1}{9})

y = - \frac{2}{27} x + \frac{1}{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}}

(3, \frac{1}{9}) : m = - \frac{2}{27}

Find the line with slope m=

- \frac{2}{27}

and passing through

(3, \frac{1}{9}) : y = - \frac{2}{27} x + \frac{1}{3}

y = - \frac{2}{27} x + \frac{1}{3}

\frac{d y}{d x} - \frac{9}{x} y = \frac{y ^{3}}{x ^{13}}

\frac{d}{d x} (\frac{- 7}{x})

f (x) = - x^{2} - x + 4, at x = - 1

y^{^{''}} + 5 y^{^{'}} + 6 y = - 10 t e^{2 t}

\frac{d}{d x} (\frac{x ^{6}}{2})

What is the tangent of y= 1/(x^2),(3, 1/9) ?
The tangent of y= 1/(x^2),(3, 1/9) is y=-2/27 x+1/3