What is the tangent of 4+2x+10xe^x ?

The tangent of 4+2x+10xe^x is y=(10(e^{a_{0}}a_{0}+e^{a_{0}})+2)x-10a_{0}^2e^{a_{0}}+4

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tangent of 4+2x+10xe^x

tangent of

4 + 2 x + 10 x e^{x}

y = (10 (e^{a_{0}} a_{0} + e^{a_{0}}) + 2) x - 10 a_{0}^{2} e^{a_{0}} + 4

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 4 + 2 a_{0} + 10 a_{0} e^{a_{0}})

Find the slope of

y = 4 + 2 x + 10 x e^{x} : \frac{d y}{d x} = 10 (e^{x} x + e^{x}) + 2

Compute the slope of

y = 4 + 2 x + 10 x e^{x}

x = a_{0} : m = 10 (e^{a_{0}} a_{0} + e^{a_{0}}) + 2

Find the line with slope m=

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

and passing through

(a_{0}, 4 + 2 a_{0} + 10 a_{0} e^{a_{0}}) : y = (10 (e^{a_{0}} a_{0} + e^{a_{0}}) + 2) x - 10 a_{0}^{2} e^{a_{0}} + 4

y = (10 (e^{a_{0}} a_{0} + e^{a_{0}}) + 2) x - 10 a_{0}^{2} e^{a_{0}} + 4

f (x) = (3 - 5 x e^{x}) (3 x + 2)

\int_{1}^{5} (x^{2} - x + 1) d x

\frac{d}{d x} (- sin (4 x))

y = e^{x^{4}}

\frac{d y}{d x} = (1 + e^{2 x})

What is the tangent of 4+2x+10xe^x ?
The tangent of 4+2x+10xe^x is y=(10(e^{a_{0}}a_{0}+e^{a_{0}})+2)x-10a_{0}^2e^{a_{0}}+4