What is the tangent of 1/x ?

The tangent of 1/x is y=-1/(a_{0)^2}x+2/(a_{0)}

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tangent of 1/x

tangent of

\frac{1}{x}

y = - \frac{1}{a _{0}^{2}} x + \frac{2}{a _{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{a _{0}})

Find the slope of

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

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 = - \frac{1}{a _{0}^{2}}

Find the line with slope m=

- \frac{1}{a _{0}^{2}}

and passing through

(a_{0}, \frac{1}{a _{0}}) : y = - \frac{1}{a _{0}^{2}} x + \frac{2}{a _{0}}

y = - \frac{1}{a _{0}^{2}} x + \frac{2}{a _{0}}

\frac{d x}{d t} = 6 - \frac{3 x}{500}

\frac{( 6 s - 4 )}{s ^{2} - 4 s + 20}

\frac{\partial}{\partial x} (x^{\frac{2}{3}})

x \to 0 + lim (e^{x} + ln (x))

\int \frac{18}{( x + 3 ) ( x ^{2} - 9 )} d x