What is the tangent of f(x)=sqrt(x)(x+1),\at x=1 ?

The tangent of f(x)=sqrt(x)(x+1),\at x=1 is y=2x

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(x)(x+1),\at x=1

tangent of

f (x) = x (x + 1), at x = 1

y = 2 x

Solution steps

Find the tangent point:

(1, 2)

Find the slope of

f (x) = x (x + 1) : \frac{df ( x )}{d x} = \frac{3 x + 1}{2 x}

Compute the slope of

f (x) = x (x + 1)

x = 1 : m = 2

Find the line with slope m=

2

and passing through

(1, 2) : y = 2 x

y = 2 x

\frac{( 8 x ^{2} + 2 x + 8 )}{x}

(1 + x) (\frac{d y}{d x}) - x y = x + x^{2}

\frac{\partial}{\partial x} (e^{2 x^{2} - 4 y^{2}})

\frac{5}{1 + x ^{2}}

\frac{d}{d t} (\frac{sin ( t )}{1 + t ^{2}})

What is the tangent of f(x)=sqrt(x)(x+1),\at x=1 ?
The tangent of f(x)=sqrt(x)(x+1),\at x=1 is y=2x