What is the tangent of f(x)=\sqrt[4]{4x+2},\at x=1 ?

The tangent of f(x)=\sqrt[4]{4x+2},\at x=1 is y= 1/(6^{3/4)}x+\sqrt[4]{6}-1/(6^{3/4)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=\sqrt[4]{4x+2},\at x=1

tangent of

Solution steps

Find the tangent point:

Find the slope of

Compute the slope of

x = 1 : m = \frac{1}{6 ^{\frac{3}{4}}}

Find the line with slope m=

\frac{1}{6 ^{\frac{3}{4}}}

and passing through

\int \frac{1}{x ^{2} 4 x ^{2} + 49} d x

\int \frac{- x ^{2} - 1}{x ^{2} - 1} d x

9 e^{x} (cos (x) - sin (x))

n = 0 \sum \infty (- x)^{n}

x \to 0 + lim (\frac{x}{ln ( \frac{1}{x} )})

What is the tangent of f(x)=\sqrt[4]{4x+2},\at x=1 ?
The tangent of f(x)=\sqrt[4]{4x+2},\at x=1 is y= 1/(6^{3/4)}x+\sqrt[4]{6}-1/(6^{3/4)}