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

  1. Feb 7, 2010 #1
    Consider this equation.

    [tex]x^3 = x + 6[/tex]

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

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

    [tex]3x^2 = 1[/tex]

    If we now put x=2, we get

    [tex]3(2)^2 = 12 = 1[/tex]

    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. jcsd
  3. Feb 7, 2010 #2
    The two functions aren't tangent to each other. They just intersect.
  4. Feb 7, 2010 #3


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    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 [itex](a-\delta, a + \delta)[/itex], 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
  5. Feb 8, 2010 #4
    So you mean that if f(x)=g(x) for all x, only then we can differentiate both sides?
  6. Feb 8, 2010 #5
    I think if you were to sketch both [tex] x^3 [/tex] and [tex] x + 6 [/tex] and their derivatives then all your questions would be answered.
  7. Feb 8, 2010 #6


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    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
  8. Feb 8, 2010 #7


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    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)".
  9. Feb 9, 2010 #8


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    Of course, good answers are already given. For an "extreme" example:

    x = 1-x



    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.
  10. Feb 10, 2010 #9


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    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.
  11. Feb 10, 2010 #10


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    I'm not quite sure why you want to repeat that. I was just providing extra geometric explanation.
  12. Feb 10, 2010 #11
    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.
  13. Feb 10, 2010 #12
    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?

    [tex]x^3 = x + 6[/tex]

    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 :)
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