MHB Derivative Problem: when does f(x)=-f'(x)

[SGR]

What function equals the negative derivative of itself?
f(x) = -f'(x)

 

[MarkFL]

Re: Derrivative Problem: when does f(x)=f'(x)

Hello, and welcome to MHB! (Wave)

What equation equals the negative derivative of its self?
f(x) = f'(x)

Do you mean:

$$f(x)=-f'(x)$$ ?

 

[SGR]

Re: Derrivative Problem: when does f(x)=f'(x)

YES! when does f(x) = -f'(x)??

 

[MarkFL]

Re: Derrivative Problem: when does f(x)=f'(x)

YES! when does f(x) = -f'(x)??

Let's use the notation of Leibniz and write:

$$\d{f}{x}=-f(x)$$

Now what if we separate variables and write:

$$\frac{1}{f(x)}\,df=-dx$$

We have divided by $f(x)$, and in doing so potentially eliminated the trivial solution $f(x)=0$, and so we need to be mindful of this in case the non-trivial solution does not include this.

What do you get when you integrate both sides of the above equation?

 

[SGR]

Re: Derrivative Problem: when does f(x)=f'(x)

Let's use the notation of Leibniz and write:

$$\d{f}{x}=-f(x)$$

Now what if we separate variables and write:

$$\frac{1}{f(x)}\,df=-dx$$

We have divided by $f(x)$, and in doing so potentially eliminated the trivial solution $f(x)=0$, and so we need to be mindful of this in case the non-trivial solution does not include this.

What do you get when you integrate both sides of the above equation?

I don't know how to integrate.

 

[MarkFL]

Re: Derrivative Problem: when does f(x)=f'(x)

I don't know how to integrate.

You haven't studied integration yet?

 

[Greg]

Re: Derivative Problem: when does f(x)=f'(x)

Hi SGR and welcome to MHB.

Do you know how to differentiate and that $e^x$ is its own derivative?

If so, consider

$$y=e^{-x}$$

 

[MarkFL]

Re: Derrivative Problem: when does f(x)=f'(x)

Let's use the notation of Leibniz and write:

$$\d{f}{x}=-f(x)$$

Now what if we separate variables and write:

$$\frac{1}{f(x)}\,df=-dx$$

We have divided by $f(x)$, and in doing so potentially eliminated the trivial solution $f(x)=0$, and so we need to be mindful of this in case the non-trivial solution does not include this.

What do you get when you integrate both sides of the above equation?

If we integrate, we get:

$$\ln|f(x)|=C-x$$

This implies:

$$f(x)=\pm e^{C-x}=\pm e^Ce^{-x}$$

Now, let $$c_1=\pm e^{C}$$ and also let $c_1=0$ (reintroducing the trivial solution we lost when separating variables) and we have $c_1\in\mathbb{R}$:

$$f(x)=c_1e^{-x}$$