What is the tangent of y=9+8x+5xe^x ?

The tangent of y=9+8x+5xe^x is y=(5(e^{a_{0}}a_{0}+e^{a_{0}})+8)x-5a_{0}^2e^{a_{0}}+9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=9+8x+5xe^x

tangent of

y = 9 + 8 x + 5 x e^{x}

y = (5 (e^{a_{0}} a_{0} + e^{a_{0}}) + 8) x - 5 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 + 8 a_{0} + 5 a_{0} e^{a_{0}})

Find the slope of

y = 9 + 8 x + 5 x e^{x} : \frac{d y}{d x} = 5 (e^{x} x + e^{x}) + 8

Compute the slope of

y = 9 + 8 x + 5 x e^{x}

x = a_{0} : m = 5 (e^{a_{0}} a_{0} + e^{a_{0}}) + 8

Find the line with slope m=

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

and passing through

(a_{0}, 9 + 8 a_{0} + 5 a_{0} e^{a_{0}}) : y = (5 (e^{a_{0}} a_{0} + e^{a_{0}}) + 8) x - 5 a_{0}^{2} e^{a_{0}} + 9

y = (5 (e^{a_{0}} a_{0} + e^{a_{0}}) + 8) x - 5 a_{0}^{2} e^{a_{0}} + 9

\int 6 - ∣ x ∣ d x

y = - x^{2} + 16, y = \frac{- 1}{4} x - 1, y = x - 6.6

y = \frac{7 x ^{2} + 5}{x ^{2} + 9}

f (x) = x e^{x}

What is the tangent of y=9+8x+5xe^x ?
The tangent of y=9+8x+5xe^x is y=(5(e^{a_{0}}a_{0}+e^{a_{0}})+8)x-5a_{0}^2e^{a_{0}}+9