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

The tangent of 1/(x^2),\at x=2 is y=-1/4 x+3/4

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

tangent of

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

y = - \frac{1}{4} x + \frac{3}{4}

Solution steps

Find the tangent point:

(2, \frac{1}{4})

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}}

x = 2 : m = - \frac{1}{4}

Find the line with slope m=

- \frac{1}{4}

and passing through

(2, \frac{1}{4}) : y = - \frac{1}{4} x + \frac{3}{4}

y = - \frac{1}{4} x + \frac{3}{4}

\frac{d}{d x} (\frac{cos ( 2 x )}{sin ( 4 x )})

f (x) = x^{3} + x^{2} + x + 5

x \to 0 - lim (\frac{1}{1 - 2 ^{\frac{1}{x}}})

\int 9 x - x d x

n = 0 \sum \infty n (\frac{1}{e})^{n}

What is the tangent of 1/(x^2),\at x=2 ?
The tangent of 1/(x^2),\at x=2 is y=-1/4 x+3/4