Derivative defined as a limit

[courteous]

Derivative is a limit of a differential quotient, that is $\displaystyle f'(a)=\lim_{h\to 0}{f(a+h)-f(a)\over h}$

Now, why is $\displaystyle (\frac{1}{f(x)})'=\lim_{h\to0}\frac{1}{h}[\frac{1}{f(x+h)}-\frac{1}{f(x)}]$ :?:

 

[galactus]

Use the quotient rule:

Let's assume f is differentiable at x and $\displaystyle f(x)\neq 0$, then $\displaystyle \frac{1}{f(x)}$ is differentiable at x and

$\displaystyle \frac{d}{dx}\left[\frac{1}{f(x)}\right]=\frac{f(x)(0)-(1)f'(x)}{[f(x)]^{2}}=\frac{-f'(x)}{[f(x)]^{2}}=\frac{d}{dx}\left[\frac{1}{f(x)}\right]$

 

[courteous]

What if you pretend you do not know quotient rule?

0)Preeminently, derivative is defined as a limit of a differential quotient: $\displaystyle \lim_{h\to 0}{f(a+h)-f(a)\over h}$.
1)Then, textbook proves (using above definition all the way): $\displaystyle [(f+g)(x)]'=f'(x)+g'(x)$.
2)Then, it proves the product rule.
3)Then, it builds on previous two ("sum rule" and product rule; still using the definition) proving quotient rule: $\displaystyle (\frac{1}{f(x)})'=\lim_{h\to0}\frac{1}{h}[\frac{1}{f(x+h)}-\frac{1}{f(x)}]=\lim_{h\to0}\frac{1}{h}[\frac{f(x)-f(x+h)}{f(x+h)f(x)}]=...=-\frac{f'(x)}{f^2(x)}$

:!: Oblivious of quotient rule, how would you show that $\displaystyle (\frac{1}{f(x)})'=\lim_{h\to0}\frac{1}{h}[\frac{1}{f(x+h)}-\frac{1}{f(x)}]=...$ :?:

 

[Deleted member 4993]

Or you can think in the following way:

Let

$\displaystyle \frac{1}{f(x)} \, = \, G(x)$

$\displaystyle \frac{1}{f(x+h)} \, = \, G(x+h)$

$\displaystyle [\frac{1}{f(x)}]' \, = \, G'(x) \, = \, \lim_{h\to 0}\frac{G(x+h) \, - \, G(x)}{h} \, = \, \lim_{h\to 0}\frac{\frac{1}{f(x+h)} \, - \, \frac{1}{f(x)}}{h}$

No quotient rule needed...

 

[galactus]

There ya' go. SK cam through with something easy.

 

[courteous]

Thank you both. I am (was) blind as bat :x . Now I'm content with underlying principles. Integrals, watch out! Thank you!