<s>[INST] 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
 [/INST] $$
d/dx \ (2x^3-5x^2+x)^{(\frac{1}{3})}
$$
let $f(x)$ be $2x^3-5x^2+x$ and $g(x)$ be $x^{(\frac{1}{3})}$
$$
d/dx \ g(f(x)) = g'(f(x))*f'(x)
$$
so
$$
d/dx \ (2x^3-5x^2+x)^{(\frac{1}{3})} = \frac{1}{3}(2x^3-5x^2+x)^{\frac{-2}{3}}*(6x^2-10x+1)
$$
which can be simplified further if need to.
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