Question

Let \(f\) be a differentiable function of period \(p\). (a) Is the function \(f^{\prime}\) periodic? Verify your answer. (b) Consider the function \(g(x)=f(2 x)\). Is the function \(g^{\prime}(x)\) periodic? Verify your answer.

Short answer

For part (a), \(f^{\prime}\) is a periodic function with the same period as \(f\), i.e., period \(p\). For part (b), if \(g(x)\) is defined as \(f(2x)\), the derivative \(g^{\prime}(x)\) is also periodic but with half the period, i.e., period \(p/2\).

Step-by-step solution

Part A: Is the Function \(f^{\prime}\) periodic?

A function is said to be periodic if it repeats its values at regular intervals or periods. Considering function \(f\) is differentiable and periodic with period \(p\), the derivative function \(f^{\prime}\) would also be periodic with the same period \(p\). This is because the nature of the slope of the function repeats after each interval of \(p\). Therefore, \(f^{\prime}\) is periodic with period \(p\). This can be formally proved by considering any point \(x\) in the domain of \(f\). As \(f\) is periodic, \(f(x)=f(x+p)\). Differentiating both sides, we get \(f^{\prime}(x) = f^{\prime}(x+p)\). So, \(f^{\prime}\) is also periodic with period \(p\).

Part B: Deriving the Function \(g(x)=f(2x)\)

First, differentiate \(g(x)=f(2x)\) using the chain rule which states that the derivative of a composite function is the derivative of the function times the derivative of the inner function. Applying the rule, the derivative \(g^{\prime}(x)\) becomes \(f^{\prime}(2x) \cdot 2\).

Part B: Is the Function \(g^{\prime}(x)\) periodic?

Considering the behaviour of the derivative of \(f\) at \(2x\), \(g^{\prime}(x)\) would also be periodic as the slope of \(f\) repeated at intervals of \(p\) will also occur for \(2x\), hence at intervals of \(p/2\). However, this is under the condition of \(x\) being multiplied by the derivative \(f^{\prime}(2x)\) and not being added a constant. Therefore, the function \(g^{\prime}(x)\) is periodic with period \(p/2\). Again, this can be formally proved by considering any point \(x\) in the domain of \(f\). As \(f(2x)\) is periodic with period \(p/2\), \(f(2x)=f(2(x+p/2))\). Differentiating both sides, we get \(f^{\prime}(2x) \cdot 2 = f^{\prime}(2(x+p/2)) \cdot 2\). So, \(g^{\prime}\) is also periodic with period \(p/2\).