What is the tangent of f(x)=x^3ln(2x),\at x=1 ?

The tangent of f(x)=x^3ln(2x),\at x=1 is y=(3ln(2)+1)x-2ln(2)-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)=x^3ln(2x),\at x=1

tangent of

f (x) = x^{3} ln (2 x), at x = 1

y = (3 ln (2) + 1) x - 2 ln (2) - 1

Solution steps

Find the tangent point:

(1, ln (2))

Find the slope of

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

Compute the slope of

f (x) = x^{3} ln (2 x)

x = 1 : m = 3 ln (2) + 1

Find the line with slope m=

3 ln (2) + 1

and passing through

(1, ln (2)) : y = (3 ln (2) + 1) x - 2 ln (2) - 1

y = (3 ln (2) + 1) x - 2 ln (2) - 1

\int_{1}^{9} y ln (y) d y

\frac{\partial}{\partial x} (\frac{x}{sin ( 2 x + 3 y + z ^{2} )})

6 4 x^{2} + 3

\frac{\partial}{\partial u} (3 v cos (u))

\frac{d}{d x} (ln ((x^{2} + 4)^{5}))

What is the tangent of f(x)=x^3ln(2x),\at x=1 ?
The tangent of f(x)=x^3ln(2x),\at x=1 is y=(3ln(2)+1)x-2ln(2)-1