What is the tangent of f(x)=((e^{2x}+1))/(e^x-2) ?

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

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tangent of f(x)=((e^{2x}+1))/(e^x-2)

tangent of

f (x) = \frac{( e ^{2 x} + 1 )}{e ^{x} - 2}

y = \frac{- e ^{a_{0}} + e ^{3 a_{0}} - 4 e ^{2 a_{0}}}{( e ^{a_{0}} - 2 ) ^{2}} x + \frac{e ^{2 a_{0}} + 1}{e ^{a_{0}} - 2} - \frac{a _{0} ( - e ^{a_{0}} + e ^{3 a_{0}} - 4 e ^{2 a_{0}} )}{( e ^{a_{0}} - 2 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{e ^{2 a_{0}} + 1}{e ^{a_{0}} - 2})

Find the slope of

f (x) = \frac{e ^{2 x} + 1}{e ^{x} - 2} : \frac{df ( x )}{d x} = \frac{- e ^{x} + e ^{3 x} - 4 e ^{2 x}}{( e ^{x} - 2 ) ^{2}}

Compute the slope of

f (x) = \frac{e ^{2 x} + 1}{e ^{x} - 2}

x = a_{0} : m = \frac{- e ^{a_{0}} + e ^{3 a_{0}} - 4 e ^{2 a_{0}}}{( e ^{a_{0}} - 2 ) ^{2}}

Find the line with slope m=

\frac{- e ^{a_{0}} + e ^{3 a_{0}} - 4 e ^{2 a_{0}}}{( e ^{a_{0}} - 2 ) ^{2}}

and passing through

(a_{0}, \frac{e ^{2 a_{0}} + 1}{e ^{a_{0}} - 2}) : y = \frac{- e ^{a_{0}} + e ^{3 a_{0}} - 4 e ^{2 a_{0}}}{( e ^{a_{0}} - 2 ) ^{2}} x + \frac{e ^{2 a_{0}} + 1}{e ^{a_{0}} - 2} - \frac{a _{0} ( - e ^{a_{0}} + e ^{3 a_{0}} - 4 e ^{2 a_{0}} )}{( e ^{a_{0}} - 2 ) ^{2}}

y = \frac{- e ^{a_{0}} + e ^{3 a_{0}} - 4 e ^{2 a_{0}}}{( e ^{a_{0}} - 2 ) ^{2}} x + \frac{e ^{2 a_{0}} + 1}{e ^{a_{0}} - 2} - \frac{a _{0} ( - e ^{a_{0}} + e ^{3 a_{0}} - 4 e ^{2 a_{0}} )}{( e ^{a_{0}} - 2 ) ^{2}}

x^{4} - 3 x^{3} + 5

y = e^{\frac{8}{9} x}

\int t^{2} e^{- 12 t} d t

\int \frac{x ^{2} + x - 6}{x ^{2} - 4 x + 3} d x

\frac{\partial}{\partial x} (\frac{4 x + y}{x - y})

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