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Lets say that I have an equation that can't really be solved via elementary means, for e.g:

$$ e^x = 4x$$

Logically, what is wrong with me using equating derivatives (or integrals for that matter)? For e.g:

$$ \dfrac{d}{dx} (e^x) = \dfrac{d}{dx} (4x) $$

or

$$ \int{e^x}{dx} = \int{4x}{dx} $$

John Tan
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    Your question is a little vague, but this is probably relevant. – Git Gud Mar 05 '15 at 01:34
  • @GitGud,what do you mean by little vague? – abel Mar 05 '15 at 01:35
  • @abel I mean exactly what I said. I don't know what kind of answer you expected. I will, however, explain why I said the question is vague. Obviously, there is nothing wrong with differentiating differentiable functions, it just might not help solving the equation. But the OP never mentioned why differentiating or integrating should be related to solving the equation, so I find the question very mysterious. If you compare it to the question I linked you'll see that the linked question does not have these problems. – Git Gud Mar 05 '15 at 01:50
  • That is a common procedure in order to solve also differential equations. This creates redundant solutions one has to eliminate. So while they solve the derived equation not all of them actually solve the original equation. – C-star-W-star Mar 05 '15 at 01:55
  • @GitGud Hi to clarify, what I am interested in is whether differentiating/integrating both sides of an equation will always lead to a correct solution. – John Tan Mar 05 '15 at 02:07
  • FWIW, the approximate value of the only real solution for the example equation is 0.3574029561813889030688111; the Inverse Symbolic Calculator finds no simple closed form for this number. – PM 2Ring Mar 05 '15 at 07:05

3 Answers3

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Interpreting both sides as numbers $$ e^x = 4x $$ the equal sign might hold for certain values of $x$.

Interpreting both sides as functions $$ \exp = 4 \mbox{ id} $$ the equal sign is not valid. The differentiation operator acts on functions, that is why the second interpretation has to be used, and explains why the results stay different as well.

If one uses a valid function equality, like $$ \tan x = \sin x / \cos x $$ then the differentiated equation is valid as well: $$ 1+ \tan^2 x = (\cos^2 x + \sin^2 x)/ \cos^2 x $$

mvw
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  • +1, this is a much clearer explanation than the accepted answer. – goblin GONE Mar 05 '15 at 11:53
  • Would an acceptable summary be that the differentiation operator preserves identical equality, but not necessarily point equality? – wchargin Mar 05 '15 at 15:16
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    @WChargin, kind of. It would be better to say: "differentiation preserves equality, but not equalizers." (The equalizer of two functions $g,f : X \rightarrow Y$ is the set of all $x \in X$ satisfying $g(x)=f(x)$.) – goblin GONE Mar 07 '15 at 00:05
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    @goblin cool, I like that. An "equalizer" is an interesting concept. – wchargin Mar 07 '15 at 00:32
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For the equations you write to be true, $x$ has to take on some discrete value. The equality is not true in general. For the equals sign to hold, both sides of the equation need to be constants because it's only an equality when x takes on specific values. This means you can't just take the derivative or integral of both sides because you're changing the nature of the function.

Take a simple example, $x^2=2x$. This has solutions $x=0$ and $x=2$. Take the derivative of both sides, and you get $2x=2$. The solutions to the second equation have nothing to do with the solutions to the first equation, so taking a derivative is not a valid approach in general when only specific solutions exist.

MonkeysUncle
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    Or, as user86418 writes here, "[w]hen you differentiate a function $f$ at one point $a$, you implicitly use the values of $f$ in some neighborhood of $a$. Since your $f$ and $g$ are not equal in any open neighborhood, you can't expect differentiating to yield a new true condition." – wchargin Mar 05 '15 at 03:38
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Logically derivatives represent the rate of change of the function,the rate of change doesn't take into consideration from what value the function starts to change at that rate.

A simple example is $2x+15=2x$.

Now the other way around if two functions take the same value at one point they don't have to have the same change rate.Other then that you can think it geometrically,they don't have to approach that point from the same path,simple example is $x$ and $-x$ the first one approaches $0$ from left and the second one from right.

And an algebraic approach that was intuitive for me if $f(x)=g(x)$ for all $x\in(c-\epsilon,c+\epsilon)$ and for any $\epsilon$ then $f'(c)=g'(c)$,it seemed clear to me that in most cases $f(x)\not=g(x)$ for all $x\in(c-\epsilon,c+\epsilon)$ so in that cases $f'(c)\not=g'(c)$

kingW3
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