Log Derivative Question

[Jason76]

Using the "log chain rule":

\(\displaystyle ln(4x)\)

\(\displaystyle u = 4x\)

\(\displaystyle u' = 4\)

\(\displaystyle f'(x) = \frac{1}{4x}(4)\)

\(\displaystyle f'(x) = \frac{1}{x}\)

Using the "log multiplication rule":

\(\displaystyle ln(4)(x)\)

\(\displaystyle ln(4) + ln(x)\)

\(\displaystyle 0 + \frac{1}{x}\)

Considering

\(\displaystyle 2lnx\)

The video I watched used the "log chain rule", but could you also use the "log multiplication rule"?

 

[Deleted member 4993]

Using the "log chain rule":

\(\displaystyle ln(4x)\)

\(\displaystyle u = 4x\)

\(\displaystyle u' = 4\)

\(\displaystyle f'(x) = \frac{1}{4x}(4)\)

\(\displaystyle f'(x) = \frac{1}{x}\)

Using the "log multiplication rule":

\(\displaystyle ln(4)(x)\)

\(\displaystyle ln(4) + ln(x)\)

\(\displaystyle 0 + \frac{1}{x}\)

Considering ← What does that mean

\(\displaystyle 2lnx\) ← Where did 2 come from??


The video I watched used the "log chain rule", but could you also use the "log multiplication rule"?

.

 

[JeffM]

Using the "log chain rule":

\(\displaystyle ln(4x)\)

\(\displaystyle u = 4x\)

\(\displaystyle u' = 4\)

\(\displaystyle f'(x) = \frac{1}{4x}(4)\)

\(\displaystyle f'(x) = \frac{1}{x}\)

Using the "log multiplication rule":

\(\displaystyle ln(4)(x)\)

\(\displaystyle ln(4) + ln(x)\)

\(\displaystyle 0 + \frac{1}{x}\)

Considering

\(\displaystyle 2lnx\)

The video I watched used the "log chain rule", but could you also use the "log multiplication rule"?

Yes. There frequently are different ways to attack a problem.

\(\displaystyle y = 2 * ln(x) \implies \dfrac{dy}{dx} = \left(\dfrac{d(2)}{dx} * ln(x)\right) + \left(2 * \dfrac{d(ln\{x\})}{dx}\right) = \{0 * ln(x)\} + \left(2 * \dfrac{1}{x}\right) = \dfrac{2}{x}.\)

\(\displaystyle u = x^2\ and\ y = 2 * ln(x) \implies \dfrac{du}{dx} = 2x\ and\ y = ln(x^2) = ln(u) \implies \dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dx} = \dfrac{1}{u} * 2x = \dfrac{1}{x^2} * 2x = \dfrac{2}{x}.\)

 

[lookagain]

Using the "log multiplication rule":

\(\displaystyle ln(4)(x)\) . . . . This is an incorrect form. You could type ln[(4)(x)] instead.

\(\displaystyle ln(4) + ln(x)\)

\(\displaystyle 0 + \frac{1}{x}\)

\(\displaystyle f(x) = ln(4x)\)


\(\displaystyle f(x) = ln[(4)(x)]\)


\(\displaystyle f(x) = ln(4) + ln(x)\)


\(\displaystyle f'(x) = 0 + \dfrac{1}{x}\)


\(\displaystyle \boxed{f'(x) = \dfrac{1}{x}}\)


------------------------------------------------------------

Using the "log chain rule":

\(\displaystyle ln(4x)\) f(x) = ln(4x)

d(ln(u)) = (1/u)du

\(\displaystyle u = 4x\)

\(\displaystyle u' = 4\)

\(\displaystyle f'(x) = \frac{1}{4x}(4)\)

\(\displaystyle f'(x) = \frac{1}{x}\)

-----------------------------------------------------------------


\(\displaystyle y = ln(4x)\)


\(\displaystyle e^y = e^{ln(4x)}\)


\(\displaystyle e^y = 4x\)


\(\displaystyle e^y\bigg(\dfrac{dy}{dx}\bigg) = 4\)


\(\displaystyle \dfrac{dy}{dx} = \dfrac{4}{e^y}\)


\(\displaystyle \dfrac{dy}{dx} = \dfrac{4}{4x}\)


\(\displaystyle \dfrac{dy}{dx} = \dfrac{1}{x}\)

 

[HallsofIvy]

Using the "log chain rule":

\(\displaystyle ln(4x)\)

\(\displaystyle u = 4x\)

\(\displaystyle u' = 4\)

\(\displaystyle f'(x) = \frac{1}{4x}(4)\)

\(\displaystyle f'(x) = \frac{1}{x}\)

Using the "log multiplication rule":

\(\displaystyle ln(4)(x)\)

\(\displaystyle ln(4) + ln(x)\)

\(\displaystyle 0 + \frac{1}{x}\)

Considering

\(\displaystyle 2lnx\)

The video I watched used the "log chain rule", but could you also use the "log multiplication rule"?

If you were "considering" \(\displaystyle ln(x^2)\) then, yes, it can be done in either of two ways:
Let \(\displaystyle u= x^2\) so that \(\displaystyle \frac{du}{dx}= 2x\) and \(\displaystyle \frac{ln(u)}{du}= \frac{1}{u}\) so that \(\displaystyle \frac{d ln(x^2)}{dx}= \frac{ln(u)}{du}\frac{du}{dx}= \frac{1}{x^2}(2x)= \frac{2}{x}\)

or, more easily, \(\displaystyle ln(x^2)= 2ln(x)\) and \(\displaystyle \frac{dln(x^2)}{dx}= 2\frac{dln(x)}{dx}= \frac{2}{x}\).