What is the tangent of y(X)=X^2+X^{-1} ?

The tangent of y(X)=X^2+X^{-1} is y=(a_{0}^3-2)/(a_{0)^3}x+3/(a_{0)^2}

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tangent of y(X)=X^2+X^{-1}

tangent of

y = x^{2} + x^{- 1}

y = \frac{a _{0}^{3} - 2}{a _{0}^{3}} x + \frac{3}{a _{0}^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{a _{0}^{3} + 1}{a _{0}^{2}})

Find the slope of

y x = x^{2} + x^{- 1} : \frac{d y}{d x} = \frac{x ^{3} - 2}{x ^{3}}

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{a _{0}^{3} - 2}{a _{0}^{3}}

Find the line with slope m=

\frac{a _{0}^{3} - 2}{a _{0}^{3}}

and passing through

(a_{0}, \frac{a _{0}^{3} + 1}{a _{0}^{2}}) : y = \frac{a _{0}^{3} - 2}{a _{0}^{3}} x + \frac{3}{a _{0}^{2}}

y = \frac{a _{0}^{3} - 2}{a _{0}^{3}} x + \frac{3}{a _{0}^{2}}

\frac{\partial}{\partial x} (x cos (x + 2 y))

\frac{\partial}{\partial x} (ln (\frac{y}{cos ( x )}))

f (x) = 5 sec (x)

g (x) = - 8 x^{2} - x + 8

x \to 2 lim (\frac{3}{x - 2})

What is the tangent of y(X)=X^2+X^{-1} ?
The tangent of y(X)=X^2+X^{-1} is y=(a_{0}^3-2)/(a_{0)^3}x+3/(a_{0)^2}