Understanding the Derivative of ln(x^2-1)^3x

[nothing123]

just wanted to know how i would differentiate y = ln(x^2 - 1)^3x

i thought id first bring the 3x out in front and then do product rule but i don't think that yielded the right answer (i don't know the right answer btw but it didnt look right)

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

 

[Hootenanny]

just wanted to know how i would differentiate y = ln(x^2 - 1)^3x

i thought id first bring the 3x out in front and then do product rule but i don't think that yielded the right answer (i don't know the right answer btw but it didnt look right)

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

Your method is correct as is your answer. Why do you say that it doesn't look right?

 

[nothing123]

is that the most simplified form?

i graphed the original function using some graphing software and automatically found what the derivative graph is supposed to look like. i then graphed the derivative i found and they did not match (two different functions). maybe i just entered it incorrectly, the graphing software is extremely picky about order of operations, brackets and all that...

 

[StatusX]

Is that suppposed to be:

y=(\ln(x^2+1))^{3x}

? If so, its probably easiest to take the ln again, so you get:

\ln(y)=3x\ln(\ln(x^2+1))

And then differentiate both sides, using the chain rule on the RHS, and then solve for dy/dx (plug back in for y).

 

[Hootenanny]

is that the most simplified form?

i graphed the original function using some graphing software and automatically found what the derivative graph is supposed to look like. i then graphed the derivative i found and they did not match (two different functions).

I would consider that the most simplified form. Well, I differentiated by hand, then integrated to check it and arrived at the original solution. Are you sure you entered the correct function in you software? (I used the same method as you btw)

 

[nothing123]

StatusX: yes that is correct function. i doubt we had to do the method you described simply because we really never did any like that in class.

Hootenanny: hey now I've got it! i tried a few variations. i had to enter the original function like this y = ln((x^2 - 1)^(3x)) - i hope that's still the same function...

 

[StatusX]

I edited the post, but it appears it wasn't the function I thought. In that case, yes, your method and answer are correct.

 

[Hootenanny]

Hootenanny: hey now I've got it! i tried a few variations. i had to enter the original function like this y = ln((x^2 - 1)^(3x)) - i hope that's still the same function...

Yes that's the same function;

y = \ln\left|(x^2 - 1)^{3x}\right|

You have to be careful when entering functions into your computer of calculator, if in doubt use brackets and plenty of them!