What is the tangent of f(x)=ln(x),\at x=e^{-7} ?

The tangent of f(x)=ln(x),\at x=e^{-7} is y=e^7x-8

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=ln(x),\at x=e^{-7}

tangent of

f (x) = ln (x), at x = e^{- 7}

y = e^{7} x - 8

Solution steps

Find the tangent point:

(e^{- 7}, - 7)

Find the slope of

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

Compute the slope of

f (x) = ln (x)

x = e^{- 7} : m = e^{7}

Find the line with slope m=

e^{7}

and passing through

(e^{- 7}, - 7) : y = e^{7} x - 8

y = e^{7} x - 8

\int \frac{1}{( \frac{1}{3} x - 8 ) ^{5}} d x

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

\frac{\partial}{\partial x} (x e^{3 x y})

n = 1 \sum \infty \frac{5 n}{2 n + 1}

\frac{d}{d x} (18 + x)

What is the tangent of f(x)=ln(x),\at x=e^{-7} ?
The tangent of f(x)=ln(x),\at x=e^{-7} is y=e^7x-8