Question

I need help using logarithmic differentiation to find the derivative of y for this problem.
y = (x^3 + 1)^3(x - 1)^5x^3

Answer 1

\[\large y=(x^3+1)^3(x-1)^5 x^3\]Let's start by taking the natural log of both sides,\[\large \ln y = \ln\left[(x^3 + 1)^3(x - 1)^5x^3\right]\]
From here, we want to apply a bunch of different rules of logs to break apart the right side.

Answer 2

Here's our first rule:\[\large \log(ab)=\log a+ \log b\]

Answer 3

\[\large \ln y = \ln(x^3 + 1)^3+\ln(x - 1)^5+\ln x^3\]
Understand that step?
Is there any specific part you're stuck on?

Answer 4

Yeah, I have no idea what to do when I see ln or log. I get completely lost. But the rule you posted makes sense though. So I understand that.

Answer 5

Using another rule:\[\large \log (a^b)=b \log a\]This allows us to bring the exponents outside as a factor on the log.

Answer 6

\[\large \ln y = 3\ln(x^3 + 1)+5\ln(x - 1)+3\ln x\]

Answer 7

From here we would take the derivative of both sides with respect to x.
Remember what the derivative of natural log gives us? :O

Answer 8

ln x?

Answer 9

yah, whats the derivative :D

Answer 10

That was my guess, is it 1/x then?

Answer 11

oh i see :) lol
yes it is.

So we'll be using that rule to take our derivative.

Answer 12

\[\large \ln y = 3\ln(x^3 + 1)+5\ln(x - 1)+3\ln x\]

Taking the derivative of the left side gives us,\[\large \frac{1}{y}\]But since we're taking the derivative with respect to x, we have to attach the y' to it, due to the chain rule.\[\large \frac{1}{y}y'\]That's the derivative of the left side.

Answer 13

Why do you do the chain rule ln y or 1/y?

Answer 14

I'm not quite sure what you're asking.
Maybe this will help. Here is the derivative rule for natural log.
\[\large \left(\ln x\right)' \qquad=\qquad \frac{1}{x}\]
When we apply it to y, we have to multiply by y' since we're taking the derivative of a non-x variable.\[\large \left(\ln y\right)' \qquad=\qquad \frac{1}{y}y'\]

Answer 15

You wrote that the y' had to be there due to the chain rule. I'm not sure I understand how the chain rule applies when it comes to ln y. I understand that the derivative for for natural log = 1/x .
So then y with respect to x is 1/y y', I just don't see where the chain rule was used there, does that make sense?

Answer 16

Lemme write it using Leibniz notation, maybe it will make sense.
We always apply the chain rule, multiplying by the derivative of the inner function.
It just turns out that sometimes it's unnecessary to do so.

Example:\[\large \frac{d}{dx}\ln x \qquad= \frac{1}{x}\cdot \frac{d}{dx}x\]We took the derivative of the natural log, and we also have to apply the chain rule and multiply by the derivative of the inner function, x.\[\large \frac{d}{dx}\ln x \qquad= \frac{1}{x}\cdot \frac{d}{dx}x \qquad=\qquad \frac{1}{x}\cdot \frac{dx}{dx}\]
This dx/dx term has no significance, so we ignore it.


But when we apply this process to a function of y,\[\large \frac{d}{dx}\ln y \qquad=\qquad \frac{1}{y}\cdot \frac{d}{dx}y \qquad=\qquad \frac{1}{y}\cdot \frac{dy}{dx}\]

Answer 17

ok, so it's just something that always happens whether you see it or not. When I think of the chain rule, the only thing that comes to mind is something like this: