What is the tangent of ((x^2+32sqrt(x)))/8 ,\at x=4 ?

The tangent of ((x^2+32sqrt(x)))/8 ,\at x=4 is y=2x+2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of ((x^2+32sqrt(x)))/8 ,\at x=4

tangent of

\frac{( x ^{2} + 32 x )}{8}, at x = 4

y = 2 x + 2

Solution steps

Find the tangent point:

(4, 10)

Find the slope of

y = \frac{x ^{2} + 32 x}{8} : \frac{d y}{d x} = \frac{1}{8} (2 x + \frac{16}{x})

Compute the slope of

y = \frac{x ^{2} + 32 x}{8}

x = 4 : m = 2

Find the line with slope m=

2

and passing through

(4, 10) : y = 2 x + 2

y = 2 x + 2

(t + 2)^{\frac{2}{3}} (2 t^{2} - 3)^{3}

\frac{d}{d x} (f (x) \cdot \frac{1}{g ( x )})

\frac{y ^{^{'}} - e ^{- t} + 6}{y} = - 6

2 t y^{^{'}} - y = 2 t y^{3} cos (t)

ln (x e^{x^{9}})

What is the tangent of ((x^2+32sqrt(x)))/8 ,\at x=4 ?
The tangent of ((x^2+32sqrt(x)))/8 ,\at x=4 is y=2x+2