What is the tangent of-1/((x-5)^2) ?

The tangent of-1/((x-5)^2) is y= 2/((a_{0)-5)^3}x+(-3a_{0}+5)/((a_{0)-5)^3}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of-1/((x-5)^2)

tangent of

- \frac{1}{( x - 5 ) ^{2}}

y = \frac{2}{( a _{0} - 5 ) ^{3}} x + \frac{- 3 a _{0} + 5}{( a _{0} - 5 ) ^{3}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, - \frac{1}{( a _{0} - 5 ) ^{2}})

Find the slope of

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

Compute the slope of

y = - \frac{1}{( x - 5 ) ^{2}}

x = a_{0} : m = \frac{2}{( a _{0} - 5 ) ^{3}}

Find the line with slope m=

\frac{2}{( a _{0} - 5 ) ^{3}}

and passing through

(a_{0}, - \frac{1}{( a _{0} - 5 ) ^{2}}) : y = \frac{2}{( a _{0} - 5 ) ^{3}} x + \frac{- 3 a _{0} + 5}{( a _{0} - 5 ) ^{3}}

y = \frac{2}{( a _{0} - 5 ) ^{3}} x + \frac{- 3 a _{0} + 5}{( a _{0} - 5 ) ^{3}}

\frac{d}{d x} (e^{x^{0}})

\int_{1}^{3} \frac{x}{( 2 x - 1 ) ^{3}} d x

\int x ln (5 x) d x

\int \frac{1}{x ( x + 1 )} d x

\int \frac{1}{( x ^{2} + 81 ) ^{\frac{3}{2}}} d x

What is the tangent of-1/((x-5)^2) ?
The tangent of-1/((x-5)^2) is y= 2/((a_{0)-5)^3}x+(-3a_{0}+5)/((a_{0)-5)^3}