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
 A: 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
$$
f'(x)=\frac{1}{3}(2x^3-5x^2+x)^{-2/3}(6x^2-10x+1)=
\frac{1}{3}\frac{6x^2-10x+1}{\sqrt[3]{(2x^3-5x^2+x)^2}}
$$
A: Your first step is incorrect.
$^3\sqrt{2x^3- 5x^2+ x}$ is NOT the third power of a square root- it is the third root:  $(2x^3- 5x^2+ x)^{1/3}$, not the "3/2" power.
A: $$
d/dx \ (2x^3-5x^2+x)^{(\frac{1}{3})}
$$
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.
A: To be more general, consider the case where you need to differentiate a function written as $$f(x)=\Big(P(x)\Big)^a$$ logarithmic differentiation makes life simpler.
Rewrite the above as $$\log\big(f(x)\big)=a\log\big(P(x)\big)$$ and differentiate, so $$\frac {f'(x)}{f(x)}=a \,\frac {P'(x)}{P(x)}$$ so $${f'(x)}=a \,\frac {P'(x)}{P(x)}\,f(x)=a\,{P'(x)}\Big(P(x)\Big)^{a-1}$$
