Contradiction about derivatives

[Oguz Khagan]

Hi everyone! before beginning the question, sorry if it is somewhat dumb. I am new to derivatives and try to understand things.

Let's suppose f is differentiable and f(x) = a. f being a constant function; whatever number we input (1, π, e, etc., even a itself), it will always output the number a. If we want to take the derivative of f(x), we will consider a as a.x⁰ and thus f'(x) =0.a.x^-1, which is 0. So, again whatever number we input f', it will always be 0. f'(1) = 0, f'(π) = 0, and f'(a) = 0.

Now, if we decide to take the derivative of f(a), by the same logic how the derivative of f(x) = x is 1, we will consider the derivative of f(a) = a is 1. So here comes the contradiction: we previously found f'(a) = 0 and now f'(a) = 1.

My question is, why is there such negotiation, and should we consider either of them correct? If so which one and why?

 

[JeffM]

[MATH]f(x) = a.[/MATH] The pronumeral a is considered here to be a constant, not to vary as x varies. Agreed?

f(6) = a. f(-4) = a. f(10^3) = a. f(a) = a. Right?

[MATH]\therefore f(x + h) = a \implies f(x + h) - f(x) = a - a = 0 \implies \dfrac{0}{h} = 0 \implies[/MATH]
[MATH]f'(x) = \lim_{h \rightarrow 0} \dfrac{f(x + h) - f(x)}{h} = \lim_{h \rightarrow 0} 0 = 0.[/MATH]
Whenever you have a doubt when you are starting differential calculus, go back to the definition of the derivative, the limit of the Newton quotient.

[MATH]g(x) = x.[/MATH] The pronumeral is considered to be a variable.

g(6) = 6. g(-4) = - 4. g(10^3) = 10^3. g(a) = a.

The functions g(x) = x and f(x) = a are entirely different except when x = a. Agreed.

[MATH]\therefore g(x + h) = x + h \implies g(x + h) - g(x) = x + h - x = h \implies \dfrac{h}{h} = 1 \implies[/MATH]
[MATH]g'(x) = \lim_{h \rightarrow 0} \dfrac{g(x + h) - g(x)}{h} = \lim_{h \rightarrow 0} 1 = 1.[/MATH]
It is true that g(a) = a = f(a). But it is not true that g'(a) = f'(a) because it is not true that 1 = 0.

 

[Dr.Peterson]

Hi everyone! before beginning the question, sorry if it is somewhat dumb. I am new to derivatives and try to understand things.

Let's suppose f is differentiable and f(x) = a. f being a constant function; whatever number we input (1, π, e, etc., even a itself), it will always output the number a. If we want to take the derivative of f(x), we will consider a as a.x⁰ and thus f'(x) =0.a.x^-1, which is 0. So, again whatever number we input f', it will always be 0. f'(1) = 0, f'(π) = 0, and f'(a) = 0.

Now, if we decide to take the derivative of f(a), by the same logic how the derivative of f(x) = x is 1, we will consider the derivative of f(a) = a is 1. So here comes the contradiction: we previously found f'(a) = 0 and now f'(a) = 1.

My question is, why is there such negotiation, and should we consider either of them correct? If so which one and why?

There are a couple important ideas you are overlooking here.

One is that when you take a derivative, you need to be aware of what you are differentiating with respect to -- that is, what variable is changing. You are not making that distinction (in part because you are using the f' notation rather than the dy/dx notation, which makes it explicit).

Another is that "take the derivative of f(a)" fails to define what function you are differentiating. The notation f(a) denotes a number, not a function: since f has already been defined, f(a) means the value of function f at a, which in this case is just a. So you are taking the derivative of a number, which you can't do! Or, you could take it as differentiating a constant function, and you've already shown that its derivative (with respect to x, not the number a) is zero.

Function notation, and derivative notation, are slippery, and you have to keep a strong grip on them. (And the advice to defer to the definition is excellent! Mathematical definitions are designed very carefully, and are much safer to work with (though often harder) than formulas derived from them, because we often forget the details in such a formula, such as when it applies.

 

[Steven G]

Do you understand what a derivative is? It is a function, called f'(x), which tell you the slope of the tangent line of f(x) at any point.

If f(x) = x = 1x+0 (in y= mx+b form) the slope of the tangent line to any point on this curve (line) is simply m or in this case 1.

If f(x) = c, where c is some constant, then f(x) is a horizontal line. The slope of a horizontal line is 0. That is the slope of the tangent line at any point on this line is 0. So f'(x) = 0