What is the tangent of f(x)=sqrt(x^2+9),\at x=4 ?

The tangent of f(x)=sqrt(x^2+9),\at x=4 is y= 4/5 x+9/5

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=sqrt(x^2+9),\at x=4

tangent of

f (x) = x^{2} + 9, at x = 4

y = \frac{4}{5} x + \frac{9}{5}

Solution steps

Find the tangent point:

(4, 5)

Find the slope of

f (x) = x^{2} + 9 : \frac{df ( x )}{d x} = \frac{x}{x ^{2} + 9}

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{4}{5}

Find the line with slope m=

\frac{4}{5}

and passing through

(4, 5) : y = \frac{4}{5} x + \frac{9}{5}

y = \frac{4}{5} x + \frac{9}{5}

h \to 0 lim (1)

\frac{d y}{d t} (0.1) t y = 6 t

4 x, 11

x \to 4 lim (\frac{6}{( x - 4 ) ^{2}})

\frac{d y}{d x} = x y + 3 x + 2 y + 6

What is the tangent of f(x)=sqrt(x^2+9),\at x=4 ?
The tangent of f(x)=sqrt(x^2+9),\at x=4 is y= 4/5 x+9/5