What is the tangent of f(x)=4^x,\at x=0 ?

The tangent of f(x)=4^x,\at x=0 is y=2ln(2)x+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=4^x,\at x=0

tangent of

f (x) = 4^{x}, at x = 0

y = 2 ln (2) x + 1

Solution steps

Find the tangent point:

(0, 1)

Find the slope of

f (x) = 4^{x} : \frac{df ( x )}{d x} = ln (2) \cdot 2^{2 x + 1}

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 = 2 ln (2)

Find the line with slope m=

2 ln (2)

and passing through

(0, 1) : y = 2 ln (2) x + 1

y = 2 ln (2) x + 1

\frac{d}{d x} (- 3 cos (x) + 4 sin (x))

y^{^{'}} - \frac{y}{x} = - x e^{- x}

f (x) = ln (\frac{3}{2}) x^{8}

\frac{d y}{d x} = 8 x y cos (x^{2})

\int 3 x^{5} d x

What is the tangent of f(x)=4^x,\at x=0 ?
The tangent of f(x)=4^x,\at x=0 is y=2ln(2)x+1