What is the tangent of y= 1/x ?

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

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tangent of y= 1/x

tangent of

y = \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}}

\int \frac{cos ( x )}{1 + sin ( x )} d x

\int sin (2 x y) d x

\frac{\partial}{\partial z} (x^{2} yz + \frac{1}{x y z ^{2}})

\int \frac{1}{25 x ^{2} + 2} d x

2 x y y^{^{'}} = 3 y^{2} + x^{2}

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