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:
$$-\frac{1}{5x\sqrt[5]{x}}$$

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:
$$
a^{-n}=\frac{1}{a^n}\\
a^{\frac{n+m}{n}}=\sqrt[n]{a^{n+m}}=\sqrt[n]{a^n\cdot a^m}=
\sqrt[n]{a^n}\cdot\sqrt[n]{a^m}=a\sqrt[n]{a^m}
$$
