What is the tangent of 5/x ?

The tangent of 5/x is y=-5/(a_{0)^2}x+(10)/(a_{0)}

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tangent of 5/x

tangent of

\frac{5}{x}

y = - \frac{5}{a _{0}^{2}} x + \frac{10}{a _{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{5}{a _{0}})

Find the slope of

y = \frac{5}{x} : \frac{d y}{d x} = - \frac{5}{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{5}{a _{0}^{2}}

Find the line with slope m=

- \frac{5}{a _{0}^{2}}

and passing through

(a_{0}, \frac{5}{a _{0}}) : y = - \frac{5}{a _{0}^{2}} x + \frac{10}{a _{0}}

y = - \frac{5}{a _{0}^{2}} x + \frac{10}{a _{0}}

\frac{d}{d x} (3 x^{3} (x - 2))

\int_{0}^{1} x^{7} x^{4} + 1 d x

\frac{d y}{d x} = \frac{e ^{x} - 1}{e ^{y}}

\int_{1}^{4} \frac{x ^{2} - 1}{x} d x

x \to 9 lim (\frac{5 x}{x - 9})