What is the tangent of f(x)=xsqrt(x),\at x=36 ?

The tangent of f(x)=xsqrt(x),\at x=36 is y=9x-108

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=xsqrt(x),\at x=36

tangent of

f (x) = x x, at x = 36

y = 9 x - 108

Solution steps

Find the tangent point:

(36, 216)

Find the slope of

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

Compute the slope of

f (x) = x x

x = 36 : m = 9

Find the line with slope m=

9

and passing through

(36, 216) : y = 9 x - 108

y = 9 x - 108

\int \frac{sec ( x ) + cos ( x )}{5 cos ( x )} d x

\frac{1}{1 + 2 x}, 0

\frac{d}{d x} ((ln (x)) (\frac{x ^{7} + 3}{x ^{3} - 10}))

\int (3 x - 7)^{9} d x

\int \frac{1}{4 x ^{2} - 9} d x

What is the tangent of f(x)=xsqrt(x),\at x=36 ?
The tangent of f(x)=xsqrt(x),\at x=36 is y=9x-108