Question

differentiate f(u)= ln(u^2-1)^3
=(1/((u^2-1)^3)))(6u(u^2-1)^2)
=(6u(u^2-1)^2)/(u^2-1)^3
=6u/(u^2-1)^3
is that correct?

Answer 1

nope... you seem to have made an error on cancelling


you have in the 2nd line of working

\[=\frac{6u(u^2 -1)^2}{(u^2-1)^3}\]

if you remove a common factor of \[(u^2 -1)^2 \]

you would be left with

\[=\frac{6u}{u^2-1}\]

hope this helps

Answer 2

thank you so much