What is the tangent of x^2-1/x ?

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

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tangent of x^2-1/x

tangent of

x^{2} - \frac{1}{x}

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, a_{0}^{2} - \frac{1}{a _{0}})

Find the slope of

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

Compute the slope of

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

x = a_{0} : m = 2 a_{0} + \frac{1}{a _{0}^{2}}

Find the line with slope m=

2 a_{0} + \frac{1}{a _{0}^{2}}

and passing through

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

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

\int \frac{( ln ( x ) ) ^{7}}{x} d x

\int_{3}^{6} \frac{2 x - 3}{6} d x

9 y^{^{''}} + 12 y^{^{'}} - 5 y = 0

\int_{0}^{π} x \cdot sin^{2} (x) d x

y^{^{''}} - y^{^{'}} - 6 y = 0, y (0) = 1, y^{^{'}} (0) = - 1

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