What is the tangent of f(x)=e^{7x^2},\at x=2 ?

The tangent of f(x)=e^{7x^2},\at x=2 is y=28e^{28}x-55e^{28}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

tangent of f(x)=e^{7x^2},\at x=2

tangent of

f (x) = e^{7 x^{2}}, at x = 2

y = 28 e^{28} x - 55 e^{28}

Solution steps

Find the tangent point:

(2, e^{28})

Find the slope of

f (x) = e^{7 x^{2}} : \frac{df ( x )}{d x} = e^{7 x^{2}} \cdot 14 x

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 = 28 e^{28}

Find the line with slope m=

28 e^{28}

and passing through

(2, e^{28}) : y = 28 e^{28} x - 55 e^{28}

y = 28 e^{28} x - 55 e^{28}

What is the tangent of f(x)=e^{7x^2},\at x=2 ?
The tangent of f(x)=e^{7x^2},\at x=2 is y=28e^{28}x-55e^{28}