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

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

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

tangent of

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

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

f (x) = ln (x^{3}) : \frac{df ( x )}{d x} = \frac{3}{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}{a _{0}}

Find the line with slope m=

\frac{3}{a _{0}}

and passing through

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

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

3 e^{3 t}

\frac{d}{d x} (ln (\frac{4 x ^{2}}{x - 6}))

\int e^{u} sin (u) d u

\frac{s + 2}{s ^{2} + s}

\int sin^{7} (2 x) cos (2 x) d x

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