What is the tangent of y=sqrt(x),(36,6) ?

The tangent of y=sqrt(x),(36,6) is y= 1/12 x+3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of y=sqrt(x),(36,6)

tangent of

y = x, (36, 6)

y = \frac{1}{12} x + 3

Solution steps

Find the slope of

y = x : \frac{d y}{d x} = \frac{1}{2 x}

Compute the slope of

y = x

(36, 6) : m = \frac{1}{12}

Find the line with slope m=

\frac{1}{12}

and passing through

(36, 6) : y = \frac{1}{12} x + 3

y = \frac{1}{12} x + 3

\frac{d}{d x} (lo g_{13} (x))

f (x) = 8 x^{2} cot (x)

\frac{d}{d t} (\frac{t}{1 + t})

n = 1 \sum \infty \frac{1}{n e ^{n}}

\frac{d ^{6}}{d x ^{6}} (x^{- 6})

What is the tangent of y=sqrt(x),(36,6) ?
The tangent of y=sqrt(x),(36,6) is y= 1/12 x+3