# Logaritmic Differentiation

• Jimmy84
You get 2x(x^2 - 1)^2 (x+1)^3 * (1 + 3x^2 + 2x). Now, factor the terms in the parenthesis. You get 2x(x^2 - 1)^2 (x+1)^3 * (1 + x)(1+2x). Now, you can combine the terms in the parenthesis. You get 2x(x^2 - 1)^2 (x+1)^3 * (1 + x + 2x + 2x^2). Now, combine the terms in the parenthesis. You get 2x(x^2 - 1)^2 (x+1)^3 * (3x

## Homework Statement

Find the derivative of y= x^2 (x^2 - 1)^3 (x+1)^4 using logaritmic differentiation.

## The Attempt at a Solution

y= x^2 (x^2 - 1)^3 (x+1)^4 = 2 ln|x| + 3 ln |(x^2)-1|+ 4 ln |x+1|

2/x + (6x)/(x^2-1) + 4/(x+1)

Is this right and if so what should I do next?

You should start by not connecting all of the steps in your problem with an '=' sign. Do a simple example first. y=x^2. ln(y)=ln(x^2)=2*ln(x). Now differentiate. y'/y=2/x. So y'=2y/x=2x^2/x=2x. That's true. Now note that y and ln(y) are not equal. Nor is y=(ln(y))'. Take the log of both sides, then solve for y'.

Dick said:
You should start by not connecting all of the steps in your problem with an '=' sign. Do a simple example first. y=x^2. ln(y)=ln(x^2)=2*ln(x). Now differentiate. y'/y=2/x. So y'=2y/x=2x^2/x=2x. That's true. Now note that y and ln(y) are not equal. Nor is y=(ln(y))'. Take the log of both sides, then solve for y'.

My problem is in the first steps, you said that y=x^2. ln(y)=ln(x^2)=2*ln(x)
Why does ln(x^2)=2*ln(x)

I don't remember much of my algebra classes about logarithms.

I would appreciate some help with this.

It's part of the properties of logarithms..

$$log (a^x) = x * log (a)$$

This is because of the exponent property...

$$(b^x)^y=b^{xy}$$

Still I am not having the result.

I tried this:
y= x^2 (x^2 - 1)^3 (x+1)^4

Lny = 2Ln(x) + 3Ln(x^2-1) + 4Ln(x+1)

y'/y = 2/x + 6x/x^2-1 + 4/x+1

y'= 2x (x^2 - 1)^3 (x+1)^4 + 6x^3 (x^2 - 1)^2 (x+1)^4 + 4 x^2 (x^2 - 1)^3 (x+1)^3

Is this right?

Put parentheses around your denominators, ok? I.e. write y'/y = 2/x + 6x/(x^2-1) + 4/(x+1). But, yes, it's right so far. Now I think you just need to try and simplify it.

Dick said:
Put parentheses around your denominators, ok? I.e. write y'/y = 2/x + 6x/(x^2-1) + 4/(x+1). But, yes, it's right so far. Now I think you just need to try and simplify it.

How can I simplify? I can't see a way of doing that.

y'= 2x (x^2 - 1)^3 (x+1)^4 + 6x^3 (x^2 - 1)^2 (x+1)^4 + 4 x^2 (x^2 - 1)^3 (x+1)^3 is also right. Take out the common factors and try and simplify what's left.

Dick said:
y'= 2x (x^2 - 1)^3 (x+1)^4 + 6x^3 (x^2 - 1)^2 (x+1)^4 + 4 x^2 (x^2 - 1)^3 (x+1)^3 is also right. Take out the common factors and try and simplify what's left.

I guess the common factors are (x^2 - 1) and (x+1) what's left is 2x + 6x^3 + 4 x^2

I guess I could simplify by 2x
2x ( 1 + 3x^3 + 2x^2 )

If this is correct what is next? I am kind of confuse I don't see where this is going.

The point is that your answer is already right. There's nothing wrong with what you did. But you said "Still I am not having the result." I assume that means you haven't simplified it enough to get the form of the answer you want. Is that right? In that case, that's where it's going. You can immediately factor out the highest power of each factor in your expression. That would be x*(x^2-1)^2*(x+1)^3. Then try and factor what's left. What is the answer you are looking for anyway?

Dick said:
The point is that your answer is already right. There's nothing wrong with what you did. But you said "Still I am not having the result." I assume that means you haven't simplified it enough to get the form of the answer you want. Is that right? In that case, that's where it's going. You can immediately factor out the highest power of each factor in your expression. That would be x*(x^2-1)^2*(x+1)^3. Then try and factor what's left. What is the answer you are looking for anyway?

The answer I am looking for is 2x(x+1)^6 (x-1)^2 (6x^2 -2x -1)

Jimmy84 said:
The answer I am looking for is 2x(x+1)^6 (x-1)^2 (6x^2 -2x -1)

Ok, that's the same as your answer in a different form. Notice (x^2-1)=(x-1)*(x+1). Just combine and factor.

Dick said:
Ok, that's the same as your answer in a different form. Notice (x^2-1)=(x-1)*(x+1). Just combine and factor.

Let me see, I still don't get the result. From
2x (x^2 - 1)^3 (x+1)^4 + 6x^3 (x^2 - 1)^2 (x+1)^4 + 4 x^2 (x^2 - 1)^3 (x+1)^3

(x^2-1)^3 = (x-1)^1.5 (x+1)^1.5 ?

So by adding all the (x^2 - 1) terms I get (x-1)^4 (x+1)^4

Plus the other (x+1) terms in the answer (x+1)^11 + (x+1)^4 = (x+1)^15 ?

Jimmy84 said:
Let me see, I still don't get the result. From
2x (x^2 - 1)^3 (x+1)^4 + 6x^3 (x^2 - 1)^2 (x+1)^4 + 4 x^2 (x^2 - 1)^3 (x+1)^3

(x^2-1)^3 = (x-1)^1.5 (x+1)^1.5 ?

So by adding all the (x^2 - 1) terms I get (x-1)^4 (x+1)^4

Plus the other (x+1) terms in the answer (x+1)^11 + (x+1)^4 = (x+1)^15 ?

Factor! Take out the common factor first. All of the terms are divisible by x*(x^2-1)^2*(x+1)^3. Factor it out.