What is the tangent of y=3+6x+8xe^x ?

The tangent of y=3+6x+8xe^x is y=(8(e^{a_{0}}a_{0}+e^{a_{0}})+6)x-8a_{0}^2e^{a_{0}}+3

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tangent of y=3+6x+8xe^x

tangent of

y = 3 + 6 x + 8 x e^{x}

y = (8 (e^{a_{0}} a_{0} + e^{a_{0}}) + 6) x - 8 a_{0}^{2} e^{a_{0}} + 3

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 3 + 6 a_{0} + 8 a_{0} e^{a_{0}})

Find the slope of

y = 3 + 6 x + 8 x e^{x} : \frac{d y}{d x} = 8 (e^{x} x + e^{x}) + 6

Compute the slope of

y = 3 + 6 x + 8 x e^{x}

x = a_{0} : m = 8 (e^{a_{0}} a_{0} + e^{a_{0}}) + 6

Find the line with slope m=

8 (e^{a_{0}} a_{0} + e^{a_{0}}) + 6

and passing through

(a_{0}, 3 + 6 a_{0} + 8 a_{0} e^{a_{0}}) : y = (8 (e^{a_{0}} a_{0} + e^{a_{0}}) + 6) x - 8 a_{0}^{2} e^{a_{0}} + 3

y = (8 (e^{a_{0}} a_{0} + e^{a_{0}}) + 6) x - 8 a_{0}^{2} e^{a_{0}} + 3

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

h (x) = 8 x^{2} - 6

x^{0.5}

\frac{\partial}{\partial x} (2 x^{2} + 3 y - 20)

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

What is the tangent of y=3+6x+8xe^x ?
The tangent of y=3+6x+8xe^x is y=(8(e^{a_{0}}a_{0}+e^{a_{0}})+6)x-8a_{0}^2e^{a_{0}}+3