What is the tangent of 2xe^x ?

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

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tangent of 2xe^x

tangent of

2 x e^{x}

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

Find the slope of

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

Compute the slope of

y = 2 x e^{x}

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

Find the line with slope m=

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

and passing through

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

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

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

\int_{0}^{1} x 1 - x^{2} d x

\int \frac{2 u - 1}{- u + 2} d u

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

\frac{d}{d x} (e^{\frac{1}{3} x})

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