What is the tangent of y=9+3x+6xe^x ?

The tangent of y=9+3x+6xe^x is y=(6(e^{a_{0}}a_{0}+e^{a_{0}})+3)x-6a_{0}^2e^{a_{0}}+9

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tangent of y=9+3x+6xe^x

tangent of

y = 9 + 3 x + 6 x e^{x}

y = (6 (e^{a_{0}} a_{0} + e^{a_{0}}) + 3) x - 6 a_{0}^{2} e^{a_{0}} + 9

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 9 + 3 a_{0} + 6 a_{0} e^{a_{0}})

Find the slope of

y = 9 + 3 x + 6 x e^{x} : \frac{d y}{d x} = 6 (e^{x} x + e^{x}) + 3

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 = 6 (e^{a_{0}} a_{0} + e^{a_{0}}) + 3

Find the line with slope m=

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

and passing through

(a_{0}, 9 + 3 a_{0} + 6 a_{0} e^{a_{0}}) : y = (6 (e^{a_{0}} a_{0} + e^{a_{0}}) + 3) x - 6 a_{0}^{2} e^{a_{0}} + 9

y = (6 (e^{a_{0}} a_{0} + e^{a_{0}}) + 3) x - 6 a_{0}^{2} e^{a_{0}} + 9

\frac{d}{d x} (6 3 x^{2})

x \to \frac{1}{8} lim (\frac{1}{x} - x^{- 1})

\int x^{2} (x + 2) d x

\int_{0}^{1} x e^{8 x} d x

\int 7 cos (x) + 8 sin (x) d x

What is the tangent of y=9+3x+6xe^x ?
The tangent of y=9+3x+6xe^x is y=(6(e^{a_{0}}a_{0}+e^{a_{0}})+3)x-6a_{0}^2e^{a_{0}}+9