3
$\begingroup$

If I have an equation, say $x^2 + 9x-93=0$, why can't I just take the derivative on both sides to get $2x = -9$ and therefore $x = -\dfrac{9}{2}$. What is the reason behind why what I did is wrong?

$\endgroup$
4
  • 12
    $\begingroup$ Because $x^2 + 9x - 93 = 0$ does not hold for all $x$. $\endgroup$ – user296602 Dec 16 '15 at 20:45
  • 9
    $\begingroup$ How about taking the derivative once more? You'd get $1 = 0$, which perfectly makes sense. $\endgroup$ – user258700 Dec 16 '15 at 20:45
  • $\begingroup$ Related: math.stackexchange.com/questions/407822/… $\endgroup$ – user236182 Dec 16 '15 at 20:53
  • 1
    $\begingroup$ @John Ryan Notice that if you take $x=-9/2$ then you get $x^2+9x-93=(81-162-372)/4=-453/4 wich is different to zero! $\endgroup$ – Khadija Mbarki Dec 16 '15 at 20:57
4
$\begingroup$

Your equation only holds for certain values of $x$. If you have two functions, places where they are equal are not necessarily places where their derivatives are equal.

If you have an identity, though, you can take the derivative of both sides and keep it valid. For instance: $$\frac{d}{dx}(\sin^2(x) + \cos^2(x)) = \frac{d}{dx}1$$

$$2\sin x \cos x - 2 \cos x \sin x = 0$$

$$0 = 0$$

$\endgroup$
1
  • $\begingroup$ I think your explanation lacks a bit on the part of why differentiating both sides of an equality doesn't preserve the original solutions. $\endgroup$ – YoTengoUnLCD Dec 16 '15 at 21:59
0
$\begingroup$

Because the derivative is the slope of the funcion $y=x^2+9x-93$ and the slope, in general, can be $0$ in a point that has not $y=0$.

$\endgroup$
0
$\begingroup$

A couple of different reasons: First of all, the equation you give is not an identity. If it were an identity, then you could differentiate both sides and obtain another valid (if not equally useful) identity.

Even if it's not an identity, differentiating both sides could, by sheer luck, produce an equation consistent with the original if the "slope" of the LHS and RHS are equal exactly when their values are equal. For instance, $x^2 = 0$ has the single solution $x = 0$, and at that point, the slope of $x^2$ is also equal to the slope of $0$, so it turns out that differentiating both sides to get

$$ 2x = 0 $$

yields an equation consistent with the original $x^2 = 0$.

However, it's worth pointing out that even when this "works", it's not logically valid; one can't actually solve the problem this way. In your example, the slopes of the two sides are not equal when their values are, so unsurprisingly, the two equations have different solutions.

$\endgroup$
1
  • $\begingroup$ Thanks, Unknown Upvoter, for the compensatory upvote. $\endgroup$ – Brian Tung Dec 16 '15 at 22:10
-1
$\begingroup$

Differentiation needs two variables that have a mutual dependence. You have only one variable.

Note the more complete way to describe differentiation is differentiation with respect to a variable.

That last bit is what you cannot do here. $x$ is related to no other variable.

The expression $x^2+9x-93=0$ is equivalent to stating that x has two possible values ( the roots of that expression.

So it's equivalent to saying something like :

$$x=constant$$

Now it's a trap I fell into myself to say that we can differentiate this and get $1=0$, but in fact there is nothing to differentiate with respect to. All we have is a statement of equality and, again, no dependence.

Why fall into this trap ? Because we start out with a mindset that we should be differentiating something and because that sexy $x$ looks beguilingly like a variable. But it's just a no-good cheap constant under that variable make-up.

Be wary of such disguises.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.