# Is it true that the derivative of 1/(random polynomial expression) is 0?

1. Oct 29, 2009

### iamsmooth

1. The problem statement, all variables and given/known data
i have for example, questions like $f(x)=\frac{1}{x^2-4}$ and $f(x)=\frac{1}{\sqrt{4-x^2}}$

2. Relevant equations
$$\frac{1}{x}=x^{-1}$$

Derivative of a constant = 0

3. The attempt at a solution

So if I rewrite

$$f(x)=\frac{1}{x^2-4}$$

as

$$f(x)=1(x^2-4)^{-1}$$

then I derive:

$$f\prime(x)=0(x^2-4)^{-1}$$

Then 0 times anything is 0. Well I guess 0 divided by anything is 0 too so I don't need to rewrite, but I think it looks better. Anyways is this true in general for the reciprocal of any equation? My class hasn't really talked about it, just an observation I'm making.

Thanks.

2. Oct 29, 2009

### Staff: Mentor

No, not true. You are writing 1/(x^2 -4) as a product: 1*(x^2 - 4)^(-1). If you differentiate this, you have to use the product rule. In general d/dx(f(x)*g(x)) != f'(x)*g'(x). That's your error.

3. Oct 29, 2009

### iamsmooth

Oh right, I forgot product rule :(

$$f\prime(x)=0(x^2-4)^{-1} + 1(-\frac{1}{2}(4-x^2)^{-3/2}(2x)$$

So the only thing happens is we wipe out the left side, so the derivative is just:

$$f\prime(x)=-\frac{1}{2}(4-x^2)^{-3/2}(2x)$$

Hope I don't make such bad mistakes on my midterm tomorrow :(

4. Oct 29, 2009

### Staff: Mentor

It's cheaper to make them here than it would be on your midterm tomorrow!

I'm betting you won't forget about using the product rule tomorrow.