Existence of Certain Functions

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turophile

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This is a 2-part problem. I can answer part (a) but I'm stuck on part (b).

(a) Is there a (nonconstant) function whose square is its derivative?

My answer: Yes. Let f(x) = – 1/x. Then [f(x)][sup:2q3e3t87]2[/sup:2q3e3t87] = 1/x[sup:2q3e3t87]2[/sup:2q3e3t87] and f'(x) = 1/x[sup:2q3e3t87]2[/sup:2q3e3t87], so [f(x)][sup:2q3e3t87]2[/sup:2q3e3t87] = f'(x).

(b) Is there a function for which the derivative of its reciprocal is the reciprocal of its derivative?

I've played around with part (b) for a while and haven't yet stumbled onto a type of function that will behave like this. But I think the only way I can answer "no" is to come up with a proof, and I'm having trouble getting started on that as well.
 
turophile said:
This is a 2-part problem. I can answer part (a) but I'm stuck on part (b).

(a) Is there a (nonconstant) function whose square is its derivative?

My answer: Yes. Let f(x) = – 1/x. Then [f(x)][sup:1ru0dwr0]2[/sup:1ru0dwr0] = 1/x[sup:1ru0dwr0]2[/sup:1ru0dwr0] and f'(x) = 1/x[sup:1ru0dwr0]2[/sup:1ru0dwr0], so [f(x)][sup:1ru0dwr0]2[/sup:1ru0dwr0] = f'(x).

(b) Is there a function for which the derivative of its reciprocal is the reciprocal of its derivative?

I've played around with part (b) for a while and haven't yet stumbled onto a type of function that will behave like this. But I think the only way I can answer "no" is to come up with a proof, and I'm having trouble getting started on that as well.
Click to expand...

\(\displaystyle \frac{d[\frac{1}{f(x)}]}{dx} \ \ = \ \ \frac{1}{f'(x)}\)

\(\displaystyle -\frac{f'(x)}{[f(x)]^2} \ \ = \ \ \frac{1}{f'(x)}\)

now continue....
 
OK, here goes my proof:

We want to prove that there is no function such that the derivative of its reciprocal is the reciprocal of its derivative.

Assume on the contrary that there is such a function.

Then the derivative of a reciprocal of a (differentiable) function is d/dx [1/f(x)] = 1/f'(x).

And the reciprocal of the derivative of that function is - f'(x)/[f(x)][sup:2yvb9sde]2[/sup:2yvb9sde] = 1/f'(x).

So we have 1/[f'(x)] = - f'(x)/[f(x)][sup:2yvb9sde]2[/sup:2yvb9sde] ? [f(x)][sup:2yvb9sde]2[/sup:2yvb9sde] = - [f'(x)][sup:2yvb9sde]2[/sup:2yvb9sde], which is not possible since the left side of the equation will be positive and the right side will be negative (if the function is not a constant) because of the squares. Therefore there is no function such that the derivative of its reciprocal is the reciprocal of its derivative.

Does that do the trick?
 
turophile said:
OK, here goes my proof:

We want to prove that there is no function such that the derivative of its reciprocal is the reciprocal of its derivative.

Assume on the contrary that there is such a function.

Then the derivative of a reciprocal of a (differentiable) function is d/dx [1/f(x)] = 1/f'(x).

And the reciprocal of the derivative of that function is - f'(x)/[f(x)][sup:b68vehqu]2[/sup:b68vehqu] = 1/f'(x).

So we have 1/[f'(x)] = - f'(x)/[f(x)][sup:b68vehqu]2[/sup:b68vehqu] ? [f(x)][sup:b68vehqu]2[/sup:b68vehqu] = - [f'(x)][sup:b68vehqu]2[/sup:b68vehqu], which is not possible since the left side of the equation will be positive and the right side will be negative (if the function is not a constant) because of the squares. Therefore there is no function such that the derivative of its reciprocal is the reciprocal of its derivative.<<< Correct

Does that do the trick? <<< No tricks - just logic!!
Click to expand...
 
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