How is 1/f(x) differentiable at 0.

[asif zaidi]

I can prove derivative of 1/f(x) = -f'(x)/(f^2).
But how can this be differentiable at x=0 (if f(x) is not equal to 0) as my class notes claim

Thanks

 

[EnumaElish]

Say f(0) = 2. This implies 1/f(0) = 1/2 and [1/f'(0)]' = -f'(0)/4.

 

[asif zaidi]

True but let me tell you my confusion ( should have posted in detail)

I proved 1/f'(x) as follows

- given a seq {an} -> a,

1/f(an) - 1/f(a) / (an -a) -> a then this function is differentiable.

Manipulating this:

( f(a) - f(an) ) /( f(a)f(an) (an-a) =

-f'(a) / ( f(a) f(an) ).

Since an-> a, f(an)-> f(a) and thus I get -f'(a)/f^2.

But this is my problem- I say an-> a. But 'a' can be any value including 0 but yet it is give that f(x) is not equal to 0.

Hope I am making sense

Thanks

Asif

 

[d_leet]

f(x) is not equal to zero, that in no way implies that a may not be zero.

 

[ktm]

A function is differentiable at a point if the derivative exists at a point. From the derivative you got, the derivative exists at 0, as seen from just plugging in 0 and knowing that $$f(x) \neq 0$$. Thus it is differentiable at 0. That is all.

 

[EnumaElish]

Example: f(x) = 2 + x. In particular, f(0) = 2 > 0. And, f'(x) = 1 for all x.

Then, derivative of 1/f(0) = -f'(0)/f(0)^2 = -1/4.