Did I do this right
f(x) = (ln(2x^2 +1)^(1/3)
f'(x)= 1/3 * (ln(2x^2 +1))^(-2/3) * (4x/(2x^2 +1))
= 4x/(3(ln(2x^2 +1))^(2/3)*(2x^2 +1))
Posted Apr 15, 2016, 12:56 AM UTC · edited
12
Loading comments…
f(x) = (ln(2x^2 +1)^(1/3)
f'(x)= 1/3 * (ln(2x^2 +1))^(-2/3) * (4x/(2x^2 +1))
= 4x/(3(ln(2x^2 +1))^(2/3)*(2x^2 +1))
Posted Apr 15, 2016, 12:56 AM UTC · edited