What is the tangent of f(x)=(ln(x))^3 ?

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

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tangent of f(x)=(ln(x))^3

tangent of

f (x) = (ln (x))^{3}

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, ln^{3} (a_{0}))

Find the slope of

f (x) = (ln (x))^{3} : \frac{df ( x )}{d x} = \frac{3 ln ^{2} ( x )}{x}

Compute the slope of

f (x) = (ln (x))^{3}

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

Find the line with slope m=

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

and passing through

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

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

\frac{d y}{d x} - \frac{1}{x} y = e^{(\frac{y}{x})}

x \to π lim ((x - π) tan (\frac{x}{2}))

\frac{\partial}{\partial x} (x y + yz + zy)

n = 0 \sum \infty \frac{3 ^{n}}{5 ^{n}}

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