Help with Logarithmic Differentiation

In summary, the conversation is about finding the derivative of a function, y = [2x + 1]^5 * [(x^4) - 3]^6, by taking the natural log of both sides and using the chain rule. However, the result obtained by the individual differs from the answer given in the book, leading to a discussion about the correct approach and potential mistakes in the process.
  • #1
communitycoll
45
0

Homework Statement


y = [2x + 1]^5 * [(x^4) - 3]^6

Homework Equations


I take the derivative of the natural log of both sides:

(y' / y) = [(10 ln(2x + 1)^4) / (2x + 1)] + [(24x^3 ln(x^4 - 3)^5) / (x^4 - 3)]

then I multiply both sides by the original function:

y' = [((10 ln(2x + 1)^4) / (2x + 1)) + ((24x^3 ln(x^4 - 3)^5) / (x^4 - 3))] * [(2x + 1)^5 * ((x^4) - 3)^6]

My book, however, says the answer is:

y' = [(10 / (2x + 1)) + (24x^3 / (x^4 - 3))] * [(2x + 1)^5 * ((x^4) - 3)^6]

^ i.e., the same thing without the logs.

Could someone please explain what I did wrong, or where I might have gone wrong?

The Attempt at a Solution


What you see above.
 
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  • #2
communitycoll said:

Homework Statement


y = [2x + 1]^5 * [(x^4) - 3]^6

Homework Equations


I take the derivative of the natural log of both sides:

(y' / y) = [(10 ln(2x + 1)^4) / (2x + 1)] + [(24x^3 ln(x^4 - 3)^5) / (x^4 - 3)]

then I multiply both sides by the original function:

y' = [((10 ln(2x + 1)^4) / (2x + 1)) + ((24x^3 ln(x^4 - 3)^5) / (x^4 - 3))] * [(2x + 1)^5 * ((x^4) - 3)^6]

My book, however, says the answer is:

y' = [(10 / (2x + 1)) + (24x^3 / (x^4 - 3))] * [(2x + 1)^5 * ((x^4) - 3)^6]

^ i.e., the same thing without the logs.

Could someone please explain what I did wrong, or where I might have gone wrong?

The Attempt at a Solution


What you see above.
It's helpful to do the problem in steps.

What did you get for ln(y) ?

Then, what did you get for the derivative of that result?
 
  • #3
SammyS said:
It's helpful to do the problem in steps.

What did you get for ln(y) ?

Then, what did you get for the derivative of that result?

I got ln(y) for ln(y), I'm not sure I understand the first question.

The derivative of that I got:

(1 / y)(y')
 
  • #4
I think that you took your derivitives wrong.

I see that you broke it up into the sum of two logs, but remember that they are of the form ln(u^n) and not of the form ln(u)^n.
 
  • #5
Villyer said:
I think that you took your derivitives wrong.

I see that you broke it up into the sum of two logs, but remember that they are of the form ln(u^n) and not of the form ln(u)^n.

I use chain rule on each of those logs; should I not?
 
  • #6
Ah, forget I asked anything. I understand what I did wrong. Thanks anyway.
 
  • #7
communitycoll said:
I got ln(y) for ln(y), I'm not sure I understand the first question.
...

If y = [2x + 1]5 * [(x^4) - 3]6, then ln(y) = __?__ .
 
  • #8
Is there a reason you didn't apply the relationship for the derivative of the product of two functions? Did the question constrain you to use logarithmic differentiation?
 

1. What is logarithmic differentiation?

Logarithmic differentiation is a method used to find the derivative of a function that contains logarithmic terms. It involves taking the natural logarithm of both sides of the equation and using properties of logarithms to simplify the expression before taking the derivative.

2. When should I use logarithmic differentiation?

Logarithmic differentiation is useful when the function contains both exponential and logarithmic terms, or when the function is in the form of a power raised to a variable. It can also be used to find the derivative of a function that is difficult to differentiate using traditional methods.

3. How do I perform logarithmic differentiation?

To perform logarithmic differentiation, take the natural logarithm of both sides of the equation. Then, use properties of logarithms to simplify the expression. Finally, take the derivative of both sides and solve for the variable of interest.

4. Are there any drawbacks to using logarithmic differentiation?

One drawback of logarithmic differentiation is that it can be time-consuming and requires a good understanding of logarithm properties. It may also result in a more complex expression which can be difficult to simplify.

5. Can logarithmic differentiation be used to find higher derivatives?

Yes, logarithmic differentiation can be used to find higher derivatives. After finding the first derivative using this method, you can continue to take the derivative of the resulting expression to find higher order derivatives.

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