What is the tangent of 8+5x+7xe^x ?

The tangent of 8+5x+7xe^x is y=(7(e^{a_{0}}a_{0}+e^{a_{0}})+5)x-7a_{0}^2e^{a_{0}}+8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of 8+5x+7xe^x

tangent of

8 + 5 x + 7 x e^{x}

y = (7 (e^{a_{0}} a_{0} + e^{a_{0}}) + 5) x - 7 a_{0}^{2} e^{a_{0}} + 8

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, 8 + 5 a_{0} + 7 a_{0} e^{a_{0}})

Find the slope of

y = 8 + 5 x + 7 x e^{x} : \frac{d y}{d x} = 7 (e^{x} x + e^{x}) + 5

Compute the slope of

y = 8 + 5 x + 7 x e^{x}

x = a_{0} : m = 7 (e^{a_{0}} a_{0} + e^{a_{0}}) + 5

Find the line with slope m=

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

and passing through

(a_{0}, 8 + 5 a_{0} + 7 a_{0} e^{a_{0}}) : y = (7 (e^{a_{0}} a_{0} + e^{a_{0}}) + 5) x - 7 a_{0}^{2} e^{a_{0}} + 8

y = (7 (e^{a_{0}} a_{0} + e^{a_{0}}) + 5) x - 7 a_{0}^{2} e^{a_{0}} + 8

\frac{d}{d x} (x sin (9 x))

\frac{d}{d x} (tan (π \frac{x}{3}))

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

\frac{d}{d x} (2.72 + 0.75 x)

\int \frac{e ^{3 x}}{4 + e ^{3 x}} d x

What is the tangent of 8+5x+7xe^x ?
The tangent of 8+5x+7xe^x is y=(7(e^{a_{0}}a_{0}+e^{a_{0}})+5)x-7a_{0}^2e^{a_{0}}+8