Gradient of $f(x)=1/x$ at $x=2$ I've not done any math for a long time, so I have what I'm sure is a stupid question, but I can't figure out what to google to get a quick answer.
I've differentiated from: $$y = x^{-1}$$
to: $$ dy/dx = -x^{-2} $$
and I'm asked to get the gradient at x=2.
$$ gradient = -2^{-2} = \frac{1}{-2^{2}}$$
Why is the correct answer $$-\frac{1}{4}$$ and not just $$\frac{1}{4}$$ I expected that when you square a negative number it becomes positive, as per the graph of $y=x^2$ for example.
Thanks in advance!
 A: You're right that when you square a negative number you get a positive number; but here, you aren't squaring a negative number. What you're doing is squaring $2$, taking the reciprocal, and only then multiplying by $-1$. That's how you get a negative result.
Regardless of the signs, however, the fact that it's negative should make sense to you, given what the graph of $y=\frac{1}{x}$ looks like (based on the geometrical interpretation of the derivative).
A: $$-2^{-2}=-(2^ {-2})=-\frac1{2^{2}}=-\frac14.$$
On the other hand, $$(-2)^{-2}=\frac1{(-2)^{2}}=\frac14.$$

Addendum

I understand how it works when the brackets are added, but WHY are the brackets added exactly?

In the first equality above, the brackets are not so much added as they are made explicit.
The way to read $$-\color{purple}{5^{-2}}$$ is to think of the exponent $-2$ as applying to as little as possible of what is on the ground floor; in other words, the base of the exponent $(-2)$ is just $5,$ not $(-5).$
This is because the convention is that (in the absence of parentheses) an exponent/power binds more tightly to the base than a minus sign prefixing the base.
Thus, $$-\color{purple}{5^{-2}}=-(\color{purple}{5^{-2}}).$$
After all, $$7-5^{-2}\ne7+(-5)^{-2}.$$
