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})

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}

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{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})

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}}

t an g e n t y = 6 sin (x), at x = \frac{π}{6}

\frac{d}{d x} (x^{6} e^{2.5 x})

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})