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

[iamsmooth]

Homework Statement

i have for example, questions like f(x)=\frac{1}{x^2-4} and f(x)=\frac{1}{\sqrt{4-x^2}}

Homework Equations

\frac{1}{x}=x^{-1}

Derivative of a constant = 0

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.

 

[Mark44]

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.

 

[iamsmooth]

Oh right, I forgot product rule :(

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

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 :(

 

[Mark44]

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.