What is the tangent of f(x)= 1/(sqrt(5-x)),\at x=0 ?

The tangent of f(x)= 1/(sqrt(5-x)),\at x=0 is y= 1/(10sqrt(5))x+1/(sqrt(5))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)= 1/(sqrt(5-x)),\at x=0

tangent of

f (x) = \frac{1}{5 - x}, at x = 0

y = \frac{1}{10 5} x + \frac{1}{5}

Solution steps

Find the tangent point:

(0, \frac{1}{5})

Find the slope of

f (x) = \frac{1}{5 - x} : \frac{df ( x )}{d x} = \frac{1}{2 ( 5 - x ) ^{\frac{3}{2}}}

Compute the slope of

f (x) = \frac{1}{5 - x}

x = 0 : m = \frac{1}{10 5}

Find the line with slope m=

\frac{1}{10 5}

and passing through

(0, \frac{1}{5}) : y = \frac{1}{10 5} x + \frac{1}{5}

y = \frac{1}{10 5} x + \frac{1}{5}

y = x^{3} - 14 x

\frac{d}{d x} ((x^{3} - 6 x) (2 - 4 x^{3}))

\int x (b - x)^{2} d x

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

\int \frac{y}{3 + y ^{2}} d y

What is the tangent of f(x)= 1/(sqrt(5-x)),\at x=0 ?
The tangent of f(x)= 1/(sqrt(5-x)),\at x=0 is y= 1/(10sqrt(5))x+1/(sqrt(5))