What is the tangent of 5xe^x ?

The tangent of 5xe^x is y=5(e^{a_{0}}+e^{a_{0}}a_{0})x-5a_{0}^2e^{a_{0}}

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tangent of 5xe^x

tangent of

5 x e^{x}

y = 5 (e^{a_{0}} + e^{a_{0}} a_{0}) x - 5 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}, 5 a_{0} e^{a_{0}})

Find the slope of

y = 5 x e^{x} : \frac{d y}{d x} = 5 (e^{x} + e^{x} x)

Compute the slope of

y = 5 x e^{x}

x = a_{0} : m = 5 (e^{a_{0}} + e^{a_{0}} a_{0})

Find the line with slope m=

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

and passing through

(a_{0}, 5 a_{0} e^{a_{0}}) : y = 5 (e^{a_{0}} + e^{a_{0}} a_{0}) x - 5 a_{0}^{2} e^{a_{0}}

y = 5 (e^{a_{0}} + e^{a_{0}} a_{0}) x - 5 a_{0}^{2} e^{a_{0}}

\frac{2 x ^{3} - 4 x ^{2} + 3}{x}

\int_{1}^{7} \frac{5}{x} d x

x = y^{2} + 6 y, y = x

n = 0 \sum \infty x^{n + 5}

x \to 4 + lim (\frac{7}{x - 4})

What is the tangent of 5xe^x ?
The tangent of 5xe^x is y=5(e^{a_{0}}+e^{a_{0}}a_{0})x-5a_{0}^2e^{a_{0}}