What is the tangent of xe^x ?

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

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tangent of xe^x

tangent of

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

y = x e^{x} : \frac{d y}{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}}

x \to 0 lim (tan (4 x))

y^{^{''}} - 4 y^{^{'}} - 5 y = 0, y (- 1) = 3, y^{^{'}} (- 1) = 9

\frac{d}{d x} (4 e^{4 e^{x} + x})

f (x) = \frac{- x ^{2} - 4}{( x ^{2} - 4 ) ^{2}}

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

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