Using logarithmic differentiate technique, find dy/dx
y = [(x-3)³(x²+1) / (2x+5)³] raised to the power of 1/5
I don't know how to type this. It is the 5th root of all those.
You have
y = [(x-3)³(x²+1) / (2x+5)³]1/5
so
ln y = (1/5)ln([(x-3)³(x²+1) / (2x+5)³] =
(1/5){ln([(x-3)³(x²+1)) - ln(2x+5)³}=
(1/5){3ln(x-3) + ln(x²+1) - 3ln(2x+5)}
Now find dy/dx and exponentiate your result to get the final result.
I hope I read your superscripts right, they're a little small for me.