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

The tangent of 3/(x^2),\at x=1 is y=-6x+9

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tangent of 3/(x^2),\at x=1

tangent of

\frac{3}{x ^{2}}, at x = 1

y = - 6 x + 9

Solution steps

Find the tangent point:

(1, 3)

Find the slope of

y = \frac{3}{x ^{2}} : \frac{d y}{d x} = - \frac{6}{x ^{3}}

Compute the slope of

y = \frac{3}{x ^{2}}

x = 1 : m = - 6

Find the line with slope m=

- 6

and passing through

(1, 3) : y = - 6 x + 9

y = - 6 x + 9

\frac{d}{d x} (11 x)

\int_{- 1}^{1} (\frac{27}{x ^{4}} - 3) d x

\int 5 x - 7 d x

y^{^{'}} = k y \cdot (100 - y)

f (x) \frac{1}{2 x + 1}, at x = 7

What is the tangent of 3/(x^2),\at x=1 ?
The tangent of 3/(x^2),\at x=1 is y=-6x+9