What is the tangent of (1+4x)^{3/2},\at x=2 ?

The tangent of (1+4x)^{3/2},\at x=2 is y=18x-9

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of (1+4x)^{3/2},\at x=2

tangent of

(1 + 4 x)^{\frac{3}{2}}, at x = 2

y = 18 x - 9

Solution steps

Find the tangent point:

(2, 27)

Find the slope of

y = (1 + 4 x)^{\frac{3}{2}} : \frac{d y}{d x} = 6 (4 x + 1)^{\frac{1}{2}}

Compute the slope of

y = (1 + 4 x)^{\frac{3}{2}}

x = 2 : m = 18

Find the line with slope m=

18

and passing through

(2, 27) : y = 18 x - 9

y = 18 x - 9

\int \frac{e ^{x}}{25 - e ^{2 x}} d x

x \to + (- 1) lim (\frac{k x + 3}{4})

\frac{d}{d x} (a x^{2} e^{5 x})

t^{4} + 3 e^{3 t} t^{2} + e^{- t} t + 1

- \frac{4 x}{x ^{2} + 1}

What is the tangent of (1+4x)^{3/2},\at x=2 ?
The tangent of (1+4x)^{3/2},\at x=2 is y=18x-9