What is the tangent of f(x)=sqrt(x),\at x=64 ?

The tangent of f(x)=sqrt(x),\at x=64 is y= 1/16 x+4

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),\at x=64

tangent of

f (x) = x, at x = 64

y = \frac{1}{16} x + 4

Solution steps

Find the tangent point:

(64, 8)

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{1}{16}

Find the line with slope m=

\frac{1}{16}

and passing through

(64, 8) : y = \frac{1}{16} x + 4

y = \frac{1}{16} x + 4

4 x y y^{^{'}} = 4 y^{2} + 3 x x^{2} + y^{2}

\int (\frac{5}{x} + \frac{12}{x ^{5}}) d x

f (x) = x (x - 1)

\int \frac{x ^{2}}{( 49 - x ^{2} ) ^{\frac{3}{2}}} d x

\int \frac{1}{( 4 - 2 x ) ^{\frac{7}{2}}} d x

What is the tangent of f(x)=sqrt(x),\at x=64 ?
The tangent of f(x)=sqrt(x),\at x=64 is y= 1/16 x+4