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}}

Compute the slope of

y = - \frac{1}{x}

x = a_{0} : 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}}

- x^{2} - x + 3

y^{^{''}} - 12 y^{^{'}} + 85 y = 0

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

f (x) = \frac{1}{x ^{6}}

\frac{\partial}{\partial y} (y^{2} + y)