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

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

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

tangent of

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

y = - 2 x + 3

Solution steps

Find the tangent point:

(1, 1)

Find the slope of

f (x) = \frac{1}{x ^{2}} : \frac{df ( x )}{d x} = - \frac{2}{x ^{3}}

Compute the slope of

f (x) = \frac{1}{x ^{2}}

x = 1 : m = - 2

Find the line with slope m=

- 2

and passing through

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

y = - 2 x + 3

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

\frac{\partial}{\partial x} (5 e^{x^{2} - 4 y})

\frac{d}{d x} (\frac{x ^{2} + 4 x - 5}{ln ( x )})

n = 1 \sum \infty n! e^{- n}

\frac{d y}{d x} = y^{2} sin (x^{2})

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