What is the tangent of ((5ln(x)))/x ?

The tangent of ((5ln(x)))/x is y=(5(1-ln(a_{0})))/(a_{0)^2}x+(10ln(a_{0})-5)/(a_{0)}

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tangent of ((5ln(x)))/x

tangent of

\frac{( 5 ln ( x ) )}{x}

y = \frac{5 ( 1 - ln ( a _{0} ) )}{a _{0}^{2}} x + \frac{10 ln ( a _{0} ) - 5}{a _{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{5 ( 1 - ln ( a _{0} ) )}{a _{0}^{2}}

Find the line with slope m=

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

and passing through

(a_{0}, \frac{5 ln ( a _{0} )}{a _{0}}) : y = \frac{5 ( 1 - ln ( a _{0} ) )}{a _{0}^{2}} x + \frac{10 ln ( a _{0} ) - 5}{a _{0}}

y = \frac{5 ( 1 - ln ( a _{0} ) )}{a _{0}^{2}} x + \frac{10 ln ( a _{0} ) - 5}{a _{0}}

n = 1 \sum \infty 4 (\frac{1}{2})^{n}

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

\int \frac{( cos ( x ) ) ^{2}}{2} d x

\int sin^{4} (x) \cdot cos^{2} (x) d x

f (x) = e^{- x} sin (x)

What is the tangent of ((5ln(x)))/x ?
The tangent of ((5ln(x)))/x is y=(5(1-ln(a_{0})))/(a_{0)^2}x+(10ln(a_{0})-5)/(a_{0)}