Derivative operator on both sides

[pyroknife]

A basic question, not a homework problem.

Say I have the expression:
5x = 10

Can I apply the derivative operator, d/dx, to both sides?

d/dx(5x)=d/dx(10) would imply 5=0.

I thought you can apply operators to both sides of an equation. Why can't you not do it in this case?

 

[andrewkirk]

It's an ambiguity about what the equation means. Is it equating functions or is it equating values?

If we write f(x) = g(x) and mean it to be an equation of functions, then we are saying that the functions are identical. That is, we are saying
$$\forall x(f(x)=g(x))$$

Contrariwise, if we write f(x) = g(x) and mean it to be an equation of values, then we are not saying anything about functions. We are saying that there is one or more value of x for which the expression on the left, evaluated with x having that value, is equal to the value on the right, evaluated with x having that value. That is, we are saying:

$$\exists x(f(x)=g(x))$$

Usually it will be obvious from the context whether an equation is an equation of values or an equation of functions.

It is only where it is an equation of functions that you can differentiate both sides and it'll still be true. In the example given, it's clearly an equation of values, because it is not true that

$$\forall x (5x=10)$$
but it is true that
$$\exists x (5x=10)$$

 

[pyroknife]

Thank you for the detailed answer. I could not figure out how to explain this earlier.
So simply put, the variable we are differentiating with respect to has to manifest itself on both sides of the equal sign?

 

[andrewkirk]

No, that's not enough. Consider the equation $x^2=1$, which has solution $x=\pm1$. The equation is of values, not functions.
Now we can re-express it with an x on either side of the equals:

$$x=\frac{1}{x}$$
Differentiate that and you'll have an equation that says $x$ must be imaginary, which is wrong.

To validly differentiate both sides you need to know that the equation holds for all values of the variable over which you are differentiating. You can do that with identities, such as:

$$\sin2\theta=2\sin\theta\cos\theta,\ \ \sin^2\theta=1-\cos^2\theta$$

You cannot do it for equations that are written as a way of specifying which values are solutions, such as:

$$(x-1)^2=2-x,\ \ \sin\theta=\sin^2\theta$$

 

[pyroknife]

No, that's not enough. Consider the equation $x^2=1$, which has solution $x=\pm1$. The equation is of values, not functions.
Now we can re-express it with an x on either side of the equals:

$$x=\frac{1}{x}$$
Differentiate that and you'll have an equation that says $x$ must be imaginary, which is wrong.

To validly differentiate both sides you need to know that the equation holds for all values of the variable over which you are differentiating. You can do that with identities, such as:

$$\sin2\theta=2\sin\theta\cos\theta,\ \ \sin^2\theta=1-\cos^2\theta$$

You cannot do it for equations that are written as a way of specifying which values are solutions, such as:

$$(x-1)^2=2-x,\ \ \sin\theta=\sin^2\theta$$

Ahhh! I see.
So to correct what I stated earlier,
So simply put, the variable we are differentiating with respect to has to manifest itself on both sides of the equal sign and in addition, for all values of this variable, both sides of the equal sign must yield the same value?

 

[andrewkirk]

You only need the second part, not the first.
eg 2=2 is an identity that is valid for all values of any variable. We can differentiate both sides with respect to any variable and the result (0=0) will be valid.