What is the tangent of 1/x ,\at x=3 ?

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

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tangent of 1/x ,\at x=3

tangent of

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

y = - \frac{1}{9} x + \frac{2}{3}

Solution steps

Find the tangent point:

(3, \frac{1}{3})

Find the slope of

y = \frac{1}{x} : \frac{d y}{d x} = - \frac{1}{x ^{2}}

Compute the slope of

y = \frac{1}{x}

x = 3 : m = - \frac{1}{9}

Find the line with slope m=

- \frac{1}{9}

and passing through

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

y = - \frac{1}{9} x + \frac{2}{3}

5 sin^{2} (x^{2})

\int \frac{1}{x sin ^{2} ( ln ( x ) )} d x

\int_{0}^{5} \frac{5 t}{( t - 6 ) ^{2}} d t

x \to 1 lim (2 x)

\frac{d}{d x} (a x^{- 2})

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