Derivative of Function with Rational Exponents $f(x)= \sqrt[3]{2x^3-5x^2+x}$ I have a question following:
$$f(x)=\sqrt[3]{2x^3-5x^2+x}$$
Here's what I did,
$$f(x)=\sqrt[3]{2x^3-5x^2+x}
\\ = (2x^3-5x^2+x)^{3\over2}
\\\\f'(x) = {3\over 2}(2x^3-5x^2+x)^{3\over2}(6x^2-10x+1)$$
Did I do this correctly?? Because I have different answer on the answer page. Can I reduce or factor any?? Or was there any mistakes?
Thanks

No, you didn't. We can write $\sqrt[3]{t}=t^{1/3}$, not $t^{3/2}$. So
$$
f(x)=(2x^3-5x^2+x)^{1/3}
$$
and
$$
f'(x)=\frac{1}{3}(2x^3-5x^2+x)^{-2/3}(6x^2-10x+1)=
\frac{1}{3}\frac{6x^2-10x+1}{\sqrt[3]{(2x^3-5x^2+x)^2}}
$$
