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

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

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

tangent of

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

y = - 2 x + 3

Solution steps

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - 2

Find the line with slope m=

- 2

and passing through

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

y = - 2 x + 3

\frac{d}{d x} (\frac{10}{x ^{2} - 4 x + 1})

\int 4 x - 4 d x

d er i v a t i v e 2 (x - 1)

\int e^{2 y - y^{2}} d y

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

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