Question

Differentiate \(\ln\left(x^3-3\right)\) by first principles.

Answer 1

\[
\begin{align*}
&\quad\lim_{h\to0}\frac{\ln\left((x+h)^3-3\right)-\ln\left(x^3-3\right)}{h}\\
&=\lim_{h\to0}\frac{1}{h}\ln\left(\frac{(x+h)^3-3}{x^3-3}\right)\\
&=\lim_{h\to0}\frac{1}{h}\ln\left(\frac{x^3+3x^2h+3xh^2+h^3-3}{x^3-3}\right)\\
&=\lim_{h\to0}\frac{1}{h}\ln\left(1+\frac{3x^2h+3xh^2+h^3}{x^3-3}\right)
\end{align*}
\]
And I am stuck.

Answer 2

\[
h=(x^3-3)t\\
h\to0\implies t\to0\\
\begin{align*}
&\quad\lim_{t\to0}\frac{1}{\left(x^3-3\right)t}\ln\left(1+3x^2t+3x\left(x^3-3\right)t^2+\left(x^3-3\right)^2t^3\right)\\
&=\frac{1}{x^3-3}\ln\left(\lim_{t\to0}\left(1+3x^2t+3x\left(x^3-3\right)t^2+\left(x^3-3\right)^2t^3\right)^\frac{1}{t}\right)
\end{align*}
\]

Answer 3

I don't understand how to use your limit. Could you explain it? @nikvist

Answer 4

@Kainui Help!

Answer 5

Can you take the derivative with first principles of just plain ln(x) by itself? Another thing is, do you know how to derive the chain rule? If you know those derivations then that's one possible method to doing this problem since you're technically just doing a specific case of those general formulas we all know and love.

Answer 6

I know how to differentiate ln(x) from first principles but I have to look up for the proof of chain rule. I remembered that the proof for chain rule is very long and cumbersome and I am not sure if I can complete this in an exam situation.

Answer 7

We know that \(\lim_{t\to0}(1+3x^2t)^\frac{1}{t}=e^{3x^2}\) and we want \(\lim_{t\to0}\left(1+3x^2t+3x\left(x^3-3\right)t^2+\left(x^3-3\right)^2t^3\right)^\frac{1}{t}=e^{3x^2}\\\)

Answer 8

Ahh I see. Honestly at this point I don't really know since this kind of stuff just amounts to doing little "algebra tricks". But I'll try to help figure it out at least.

Answer 9

\(\lim_{t\to0}\left(1+3x^2t+3x\left(x^3-3\right)t^2+\left(x^3-3\right)^2t^3\right)^\frac{1}{t}=e^{3x^2}\) is true by Mathematica but how do you prove it?

Answer 10

@Callisto @ganeshie8 @Marki

Answer 11

\(\Large \lim \limits_{h\to0}\frac{1}{h}\ln\left(1+\frac{3x^2h+3xh^2+h^3}{x^3-3}\right)
= \lim \limits_{h\to0}\frac{1}{h}\ln\left(1+h f(x,h)\right)
\)

\( \large = \lim \limits_{h\to0} f(x,h)\frac{1}{f(x,h)h}\ln\left(1+h f(x,h)\right) \\ \Large = f(x,0) \ln \lim \limits_{h\to0}\left(1+h f(x,h)\right)^{\frac{1}{f(x,h)h}}\)

\(\Large = f(x,0) \ln e = f(x,0) \)

Answer 12

What would be the function f(x,h) ?
plug in h =0 there and you're done :)

Answer 13

lim x->0 (1+x)^(1/x) = e

A standard limit formula

Answer 14

yes!! Bernoulli interets compounding formula is the true basis for all of this...

Answer 15

\[f(x,h)=\frac{3x^2+3xh+h^2}{x^3-3}?\]

Answer 16

no thomas, that would become 1+ f(x,h)

but we need
1+ h f(x,h)

so just divide that by h