What is the tangent of y=x+e^x-1/x ?

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

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tangent of y=x+e^x-1/x

tangent of

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

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

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

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

Find the line with slope m=

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

and passing through

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

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

x^{2} + x y + 2 y^{2} = 58, (- 5, - 3)

\int y^{3} ln (y) d y

\frac{d}{d x} (91 + 2 \cdot 4)

\int \frac{2}{x ^{2}} d x

\int_{7}^{12} 1 d x

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