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

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

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tangent of 1/(x-5)

tangent of

\frac{1}{x - 5}

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Compute the slope of

y = \frac{1}{x - 5}

x = a_{0} : m = - \frac{1}{( a _{0} - 5 ) ^{2}}

Find the line with slope m=

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

and passing through

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

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

y = \frac{1}{x - 6}, (5, - 1)

\int x^{- 5} + x^{\frac{1}{2}} d x

\frac{d}{d x} ((- x^{2} + 25)^{2})

\int \frac{7}{x ^{3} - 125} d x

\frac{\partial}{\partial y} (sin^{2} (x - 3 y))

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