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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!

Prime Mover
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Jack Shone
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    $-a^2$ is not the same as $(-a)^2$. – lulu Mar 08 '22 at 19:52
  • $2^2=4$, and putting a negative sign on both sides we obtain that $-2^2=-4$. – azif00 Mar 08 '22 at 20:01
  • When you differentiated, you multiplied by $-1$, so your gradient is $(-1)\times x^{-2}$ i.e. $-x^{-2}$. This is not the same as $(-1\times x)^{-2}$, which is $(-x)^{-2}$. – A.M. Mar 08 '22 at 20:26

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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).

Kman3
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  • I'm embarrassed to ask, but why is that? Is that actually the standard approach when doing any math and I was just unaware? Thanks Kman! – Jack Shone Mar 08 '22 at 19:58
  • @JackShone Basically, yes. When you see $-x^{-2}$, you interpret it as $- \left(\dfrac 1 {x^2} \right)$ or equivalently $\dfrac 1 {-(x^2)}$, and not $\dfrac 1 {(-x)^2}$ because the minus sign goes outside the square function not inside it. – Prime Mover Mar 08 '22 at 20:06
  • @JackShone Don't be embarrassed; it's part of learning. When you take a derivative, the rule when it comes to constant powers is that you bring down the power as a coefficient. It doesn't change what number you input. So you have $-(x)^{-2}$ and not $(-x)^{-2}$ as I'm guessing you're thinking. And like others have mentioned, the standard notation is for $-x^2$ to mean $-(x)^2$ where you perform the squaring first. If someone wants you to take the negative of $x$ first, the notation is $(-x)^2$ to signify that the negative sign plays a role. – Kman3 Mar 08 '22 at 20:16
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$$-2^{-2}=-(2^ {-2})=-\frac1{2^{2}}=-\frac14.$$

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, $$7-5^{-2}\ne7+(-5)^{-2}.$$

ryang
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  • Thanks. I understand how it works when the brackets are added, but WHY are the brackets added exactly? Why is this done in that way, but y=x^2 is 4 at x=-2. It seems the same situation in my brain... I obviously overlooked/missed a basic principle somewhere I guess. – Jack Shone Mar 08 '22 at 20:07