# How are we able to differentiate both sides of an equation?

1. Feb 7, 2010

### Juwane

Consider this equation.

$$x^3 = x + 6$$

If we solve for x we'll get x=2

Now if we differentiate both sides with respect to x, we get

$$3x^2 = 1$$

If we now put x=2, we get

$$3(2)^2 = 12 = 1$$

Which is of course wrong. So how can we differentiate both sides of an equation when the value of x changes after the differentiation?

2. Feb 7, 2010

### Anonymous217

The two functions aren't tangent to each other. They just intersect.

3. Feb 7, 2010

### Hurkyl

Staff Emeritus
Why do you think
f(a) = g(a)​
should imply
f'(a) = g'(a)​
?

If these functions were actually equal -- i.e. f = g or equivalently f(x)=g(x) for all x -- then the derivatives would be equal functions too -- i.e. f' = g' or equivalently f'(x) = g'(x) -- and thus f'(a) would be equal to g'(a).

From a more geometric perspective, arithmetic operations act pointwise -- e.g. (f+g)(a) = f(a) + g(a) -- but derivatives require knowledge of the neighborhood of a point -- knowing f(a) doesn't tell you anything about f'(a).

The analogous statement for derivatives is that if f(c)=g(c) for all c in the interval $(a-\delta, a + \delta)$, then f'(a) = g'(a).

One third comment. If the premise of a problem is that x is a number such that x3=x+6, then differentiation with respect to x is nonsense.

Last edited: Feb 7, 2010
4. Feb 8, 2010

### Juwane

So you mean that if f(x)=g(x) for all x, only then we can differentiate both sides?

5. Feb 8, 2010

### monomath

I think if you were to sketch both $$x^3$$ and $$x + 6$$ and their derivatives then all your questions would be answered.

6. Feb 8, 2010

### HallsofIvy

Yes, if "F" is any kind of operation that gives a unique answer and a= b, then F(a)= F(b). That's what "unique" means!

Since the derivative is an operation of functions, if f(x)= g(x) are equal functions, then f'(x)= g'(x) for all x and, in particular, f'(a)= g'(a).

Since the derivative is NOT an operation on individual numbers, the fact that f(a)= g(a) are equal numbers does NOT mean that f'(a)= g'(a)

Last edited by a moderator: Feb 10, 2010
7. Feb 8, 2010

### Hurkyl

Staff Emeritus
Because equality of functions is pointwise: If "f(x)=g(x) for all x" holds, then "f=g" holds.

And if "f=g" holds, we can substitute: "f'" is equal to "g'".

Then we can apply the fact equality is pointwise again to get: "f'(x)=g'(x)".

Since differentiation is local, we also have that "f(x)=g(x) for all x in (a,b)" implies "f'(x)=g'(x) for all x in (a,b)".

8. Feb 9, 2010

### Landau

x = 1-x

differentiate

1=-1

If we substitute x=1/2, a solution of the first equation, we still have 1=-1, a contradiction :)

If you draw the graphs of x and 1-x, you will see that they are in fact perpendicular to each other in the intersection x=1/2! (which was already obvious from the 1 and -1 obtainend from differentiating.) So no way their derivatives are the same in that point: they would have to be parallel, but they are in fact the opposite, perpendicular.

9. Feb 10, 2010

### HallsofIvy

The point being that the original equation, x= 1- x, is only true for x= 1/2. It is NOT an equality of functions. It does NOT say that f(x)= x and g(x)= 1- x are equal for all x.

10. Feb 10, 2010

### Landau

I'm not quite sure why you want to repeat that. I was just providing extra geometric explanation.

11. Feb 10, 2010

### lugita15

Here's another way to think about it. If you have an equation like f(a)=g(a) for some particular value a, then you are absolutely justified in differentiating both sides. But you won't get f'(a)=g'(a). Instead you will get 0=0, because both f(a) and g(a) are constants. (This follows from the fact that a is a constant).

By the way, this reminds me of a riddle that a teacher once posed to me in high school: Since multiplication is repeated addition, if x is some particular natural number then x^2=x+x+x+x+...x (x times). If we differentiate both sides we get
2x=1+1+1+1+...1 (x times) = x, and therefore 1=2.
The resolution is the same: since both sides are constants, when we differentiate the equation we get 0=0.

12. Feb 10, 2010

### Tac-Tics

Think carefully about what it means to be doing what you're doing.

From the very start, you have two functions. Let's give them names for clarity:

f(x) = x^3
g(x) = x + 6

Note that f and g are not the same function. They are not equal.

What's the first line you put down?

$$x^3 = x + 6$$

Boom. You say just the opposite. That they ARE equal. That's the first misstep.

What you really meant to say wasn't that the functions are equal. In fact, it wasn't even a statement. It was more like a question. "For what values of x are f(x) and g(x) equal?" In set notation, you'd write this as {x | f(x) = g(x)}.

What you're interested in is the few values of x where f(x) = g(x) [in this case, there's only one: x = 2]. So we're working with the assumption that whatever value x is, f(x) = g(x). We can then investigate the values of x.

Now here's a subtle part that's leading you stray. You can take the derivative of both sides only if both sides are functions. This is not the case.

But f(x) is a function, right? Wrong. The function is called "f". When we write "f(x)", that's "the evaluation of f at x". They are two different things!!! If you have f = g, then, by all means, f' = g'. In fact, if f = g, then f'(x) = g'(x), because you're again doing the same thing to both sides of an equation (evaluating both sides at x).

When you say f(x) = g(x), you're talking about two numbers being equal, not two functions. And you can't take the derivative of a number. If we are working with a problem that says "f(x) = g(x) for all x", then we can conclude f = g. But this is not the case here! In this problem, we are told f(x) = g(x) for a handful of values. That's totally legit, but you can't conclude f = g and therefore you can't conclude that f' = g' nor f'(x) = g'(x).

Hope that helps a bit :)