Derivative of $4x^5 \tan(\frac{-1}{x})$ I can't seem to be able to get the correct signs for this derivative.
$\frac{d}{dx}(4x^5\tan(\frac{-1}{x}))$
Here's my work:
$= (4x^5)'\tan(\frac{-1}{x}) + 4x^5(\tan(\frac{-1}{x}))'$
$= 20x^4\tan(\frac{-1}{x}) + 4x^5(\tan(\frac{-1}{x}))'$
$= 20x^4\tan(\frac{-1}{x}) + 4x^5(\frac{-1}{x})'\sec^2(\frac{-1}{x})$
$= 20x^4\tan(\frac{-1}{x}) + 4x^5(\frac{-1}{x})'\sec^2(\frac{-1}{x})$
And here is where I think I make my mistake:
$(\frac{-1}{x})' = (-1x^{-1})'$
$(\frac{-1}{x})' = (1x^{-2})$
$(\frac{-1}{x})' = \frac{1}{x^2}$
So plugging that derivative back into the step I left off on in the last equation:
$= 20x^4tan(\frac{-1}{x}) + 4x^5\frac{1}{x^2}sec^2(\frac{-1}{x})$
$= 20x^4tan(\frac{-1}{x}) + 4x^3sec^2(\frac{-1}{x})$
However the answer key and wolfram alpha display the answer to be
$= 20x^4\tan(\frac{-1}{x}) - 4x^3\sec^2(\frac{-1}{x})$
I can't figure out the step where I went wrong... Please help!

Hint $sec(-\theta)=sec(\theta)$ thats the only mistake.
