What is the tangent of 8+2x+xe^x ?

The tangent of 8+2x+xe^x is y=(e^{a_{0}}a_{0}+e^{a_{0}}+2)x-a_{0}^2e^{a_{0}}+8

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tangent of 8+2x+xe^x

tangent of

8 + 2 x + x e^{x}

y = (e^{a_{0}} a_{0} + e^{a_{0}} + 2) x - a_{0}^{2} e^{a_{0}} + 8

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 8 + 2 a_{0} + a_{0} e^{a_{0}})

Find the slope of

y = 8 + 2 x + x e^{x} : \frac{d y}{d x} = e^{x} x + e^{x} + 2

Compute the slope of

y = 8 + 2 x + x e^{x}

x = a_{0} : m = e^{a_{0}} a_{0} + e^{a_{0}} + 2

Find the line with slope m=

e^{a_{0}} a_{0} + e^{a_{0}} + 2

and passing through

(a_{0}, 8 + 2 a_{0} + a_{0} e^{a_{0}}) : y = (e^{a_{0}} a_{0} + e^{a_{0}} + 2) x - a_{0}^{2} e^{a_{0}} + 8

y = (e^{a_{0}} a_{0} + e^{a_{0}} + 2) x - a_{0}^{2} e^{a_{0}} + 8

f (x) = \frac{4 x}{x + 3}, at x = - 5

x \to 1.7 lim (sin (x))

\int - \frac{19}{x ^{2} + 1} d x

e^{x} + e^{- x}

f (t) = sin (t)

What is the tangent of 8+2x+xe^x ?
The tangent of 8+2x+xe^x is y=(e^{a_{0}}a_{0}+e^{a_{0}}+2)x-a_{0}^2e^{a_{0}}+8