Question

Consider a function \(f(x)\) and its derivative \(d f / d x\). Does this derivative have to be a function of \(x\) ?

Short answer

Answer: No, the derivative of a function does not always have to be a function of \(x\). In the case of constant functions, the derivative is a constant value and does not depend on \(x\). However, for non-constant functions, the derivative may be a function of \(x\).

Step-by-step solution

Recall the definition of a derivative

The derivative of a function \(f(x)\) is defined as the limit of the difference quotient: \[\frac{df}{dx} = \lim_{h \to 0}\frac{f(x+h) - f(x)}{h}\]

Consider constant functions

If \(f(x)\) is a constant function, that means for any \(x\), \(f(x) = C\), where \(C\) is a constant. Therefore, the difference quotient becomes: \[\frac{f(x+h) - f(x)}{h} = \frac{C-C}{h} = 0\] As \(h\) goes to \(0\), this limit remains \(0\). That means, for constant functions, the derivative \(\frac{df}{dx}\) is equal to a constant, which is not dependent on \(x\). This suggests that the derivative does not always have to be a function of \(x\).

Consider a non-constant function example

Let's consider the function \(f(x) = x^2\). Using the definition of the derivative, we get: \[\frac{df}{dx} = \lim_{h \to 0}\frac{((x+h)^2 - x^2)}{h} = \lim_{h \to 0}\frac{(x^2 + 2xh + h^2 - x^2)}{h}\] \[= \lim_{h \to 0}(2x + h) = 2x\] In this case, the derivative \(\frac{df}{dx}\) is a function of \(x\), specifically \(2x\).

Conclusion

The derivative of a function \(\frac{df}{dx}\) is not always a function of \(x\). In the case of constant functions, the derivative is a constant value and does not depend on \(x\). However, for non-constant functions, the derivative may be a function of \(x\).