Asked by ana
y= 4th root (x^2+1)/(x^2-1)
Answers
Answered by
Reiny
y = [(x^2 + 1)(x^2-1) ]^(1/4)
= (x^4 - 1)^(1/4)
ln y = (1/4) ln(x^4 -1)
(dy/dx) / y = (1/4)(4x^3)/(x^4 - 1)
dy/dx = (1/4)(y)(4x^3)/(x^4 - 1)
or
you can replace y with (x^4 - 1)^(1/4)
and simplify that a bit since the denominator is x^4 - 1
= (x^4 - 1)^(1/4)
ln y = (1/4) ln(x^4 -1)
(dy/dx) / y = (1/4)(4x^3)/(x^4 - 1)
dy/dx = (1/4)(y)(4x^3)/(x^4 - 1)
or
you can replace y with (x^4 - 1)^(1/4)
and simplify that a bit since the denominator is x^4 - 1
Answered by
drwls
Do you want the derivative of that? Does the "fourth root" apply to the numerator only, or the complete fraction
(x^2+1)/(x^2-1) ?
You need to write the function in a clear nonambiguous manner, using parentheses where necessary. Use ^1/4 for fourth roots.
(x^2+1)/(x^2-1) ?
You need to write the function in a clear nonambiguous manner, using parentheses where necessary. Use ^1/4 for fourth roots.
Answered by
Reiny
Sorry ana, my answer is incorrect,
I read that as a multiplication , should have been a division
ln y = (1/4) (ln (x^2 + 1) - ln(x^2 -1)
(dy/dx) / y = (1/4) ( 2x/(x^2+1) - 2x/(x^2 + 1) )
dy/dx = (1/4)(y) (4x)/(x^4 - 1)
= xy/(x^4 - 1)
replace y with the original if you have to.
I read that as a multiplication , should have been a division
ln y = (1/4) (ln (x^2 + 1) - ln(x^2 -1)
(dy/dx) / y = (1/4) ( 2x/(x^2+1) - 2x/(x^2 + 1) )
dy/dx = (1/4)(y) (4x)/(x^4 - 1)
= xy/(x^4 - 1)
replace y with the original if you have to.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.