How to differentiate $\sqrt[5]{1/x}$ Hey guys I could use some help with this derivative:
$$\sqrt[5]{1/x}$$
This is what I have so far:
$$=-\dfrac{1}{5}\left(\dfrac{1}{x}\right)^{-6/5}$$
Having trouble simplifying this to my given solution so I don't know if it is correct.
Given solution:
$$-\frac{1}{5x\sqrt[5]{x}}$$
 A: Well, you're right that
$$\frac{d}{dx} x ^ n = n x ^ {n-1}$$
 and you're right that for $p$, $q$ integers, following that $x^{-1} = \frac 1x$, $x^{\frac 1q} = \sqrt[q]x$ and $(x^p)^q = x^{p\cdot q}$ you have
$$x^{-\frac pq} = \frac{1}{\sqrt[q]{x^p}}.$$
In you case that mean the derivative is $$\frac{-1}{5} x ^ {\frac{-6}5} = \frac{-1}{5 \sqrt[5]{x^6}}.$$
A: After the power and chain rule we get
$$\frac{1}{5}\left(\frac{1}{x}\right)^{1/5-1}\cdot \left(-\frac{1}{x^2}\right)$$
A: Here's my take on it with each step spelled out as clearly as possible:
$$
\left(\sqrt[5]{\frac{1}{x}}\right)'=
\left(x^{-\frac{1}{5}}\right)'=
-\frac{1}{5}x^{\left(-\frac{1}{5}-1\right)}=
-\frac{1}{5}x^{\left(-\frac{1}{5}-\frac{5}{5}\right)}=
-\frac{1}{5}x^{-\frac{6}{5}}=\\
-\frac{1}{5\sqrt[5]{x^6}}=
-\frac{1}{5\sqrt[5]{x^{5+1}}}=-\frac{1}{5\sqrt[5]{x^5\cdot x}}=
-\frac{1}{5\sqrt[5]{x^5}\cdot\sqrt[5]{x}}=
-\frac{1}{5x\sqrt[5]{x}}
$$
As you can see, it matches the provided solution on the nose.
The tricky algebra below is probably what threw you off:
$$
a^{-n}=\frac{1}{a^n}\\
a^{\frac{n+m}{n}}=\sqrt[n]{a^{n+m}}=\sqrt[n]{a^n\cdot a^m}=
\sqrt[n]{a^n}\cdot\sqrt[n]{a^m}=a\sqrt[n]{a^m}
$$
A: Since $(x^\alpha )' = \alpha x^{\alpha -1}$ and 
$$f(x) = \sqrt[5]{1/x} =\left(\frac1x\right)^{\frac15} =\left(x^{-1}\right)^{\frac15} = x^{-\frac15}$$
then $$f'(x) = -\frac15 x^{-\frac15 -1}=-\frac15 x^{-\frac65 }$$
A: Rewrite initial function as $$\sqrt[5]{\frac{1}{x}}=\sqrt[5]{x^{-1}}=x^{-1/5}$$ then
$$\frac{d}{dx}x^{-1/5}=-\frac{1}{5}x^{-\frac{6}{5}}$$
