Derivative of Function with Rational Exponents $f(x)= \sqrt[3]{2x^3-5x^2+x}$

Question

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

Answer 1

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 $$ f'(x)=\frac{1}{3}(2x^3-5x^2+x)^{-2/3}(6x^2-10x+1)= \frac{1}{3}\frac{6x^2-10x+1}{\sqrt[3]{(2x^3-5x^2+x)^2}} $$

Answer 2

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.

Answer 3

$$ d/dx \ (2x^3-5x^2+x)^{(\frac{1}{3})} $$

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.

Answer 4

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 $${f'(x)}=a \,\frac {P'(x)}{P(x)}\,f(x)=a\,{P'(x)}\Big(P(x)\Big)^{a-1}$$