What is the tangent of f(x)=sqrt(x),\at x=49 ?

The tangent of f(x)=sqrt(x),\at x=49 is y= 1/14 x+7/2

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=49

tangent of

f (x) = x, at x = 49

y = \frac{1}{14} x + \frac{7}{2}

Solution steps

Find the tangent point:

(49, 7)

Find the slope of

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

Compute the slope of

f (x) = x

x = 49 : m = \frac{1}{14}

Find the line with slope m=

\frac{1}{14}

and passing through

(49, 7) : y = \frac{1}{14} x + \frac{7}{2}

y = \frac{1}{14} x + \frac{7}{2}

\frac{e ^{x}}{x}

\frac{d y}{d x} = 1 + \frac{2}{( y - x ) e ^{3 (y - x)}}

x \to 1 lim (ln (3) + e^{x})

\frac{d}{d x} (ln (4 x - 2 y + 6))

y^{^{''}} + 9 y = 3 sin (2 x)

What is the tangent of f(x)=sqrt(x),\at x=49 ?
The tangent of f(x)=sqrt(x),\at x=49 is y= 1/14 x+7/2