What is the tangent of f(x)=xe^x ?

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

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tangent of f(x)=xe^x

tangent of

f (x) = x e^{x}

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

Find the slope of

f (x) = x e^{x} : \frac{df ( x )}{d x} = e^{x} + e^{x} x

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 = e^{a_{0}} + e^{a_{0}} a_{0}

Find the line with slope m=

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

and passing through

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

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

\int ln (7 x^{5}) d x

\frac{\partial}{\partial x} (e^{x} sin (y))

\frac{d y}{d x} - 9 y^{2} e^{- 3 x} = 0, y (0) = 10

\int (\frac{3}{2 x ^{\frac{1}{2}}} - \frac{2}{3 x ^{\frac{1}{3}}}) d x

\frac{d}{d x} (\frac{2}{3} x^{\frac{3}{2}} + 1)

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